Solved Examples: Inverse Trigonometric Functions

Example 1: Principal Solutions of Sine

Question: Find the principal solutions of $\cos x = \sqrt{3}/2$.

Explanation:

1. Identify Quadrants: $\cos x$ is positive, so solutions lie in Quadrant I and Quadrant IV.
2. Find Q1 Solution: The base angle $\alpha$ where $\cos \alpha = \sqrt{3}/2$ is $\alpha = \pi/6$. This is the first principal solution.
3. Find Q4 Solution: The rule for Q4 is $2\pi - \alpha$. So, the second solution is $x = 2\pi - \pi/6 = 11\pi/6$.

Answer: The principal solutions are $\pi/6$ and $11\pi/6$.

Example 2: Principal Solutions of Tangent

Question: Find the principal solutions of $\tan x = -1$.

Explanation:

1. Identify Quadrants: $\tan x$ is negative, so solutions are in Q2 and Q4.
2. Find Reference Angle ($\alpha$): The angle for the positive value, $\tan \alpha = 1$, gives $\alpha = \pi/4$.
3. Find Q2 Solution: The rule is $\pi - \alpha$. So, $x = \pi - \pi/4 = \mathbf{3\pi/4}$.
4. Find Q4 Solution: The rule is $2\pi - \alpha$. So, $x = 2\pi - \pi/4 = \mathbf{7\pi/4}$.

Answer: The principal solutions are $3\pi/4$ and $7\pi/4$.

Example 3: General Solution of Sine

Question: Find the general solution of $\sin x = -1/2$.

Explanation:

1. Find a Principal Value ($\alpha$): We need an angle $\alpha$ where $\sin \alpha = -1/2$. A convenient choice is the angle in Q4 within the principal branch of $\sin^{-1}$, which is $\alpha = -\pi/6$.
2. State the General Formula: The general solution for $\sin x = \sin \alpha$ is $$x = n\pi + (-1)^n \alpha.$$
3. Substitute and State Final Answer: $$x = n\pi + (-1)^n (-\pi/6) = n\pi - (-1)^n (\pi/6)$$

Answer: $x = n\pi - (-1)^n (\pi/6), n \in \mathbb{Z}$.

Example 4: General Solution of Cosine

Question: Find the general solution of $\cos(3x) = \cos(x)$.

Explanation: Using the formula $A = 2n\pi \pm B$ for $\cos A = \cos B$:

$3x = 2n\pi \pm x$.

This gives two cases:

• Case 1: $$3x = 2n\pi + x$$ $$\implies 2x = 2n\pi$$ $$\implies x=n\pi$$
• Case 2: $$3x = 2n\pi - x$$ $$\implies 4x = 2n\pi$$ $$\implies x=n\pi/2$$

The set of solutions $\{n\pi\}$ is a subset of $\{n\pi/2\}$ (when n is even).

Therefore, the complete, most general solution is the larger set.

Answer: $x=n\pi/2, n \in \mathbb{Z}$.

Example 5: General Solution of Tangent

Question: Find the general solution of $\tan(5x) = 0$.

Explanation: The general solution for $\tan\theta=0$ is $\theta=n\pi$, where $n$ is an integer. In this case, $\theta=5x$. So, $5x=n\pi$. Dividing by 5 gives the solution for x.

Answer: $x=n\pi/5, n \in \mathbb{Z}$.

Example 6: General Solution of Squared Functions

Question: Find the general solution of $\sin^2x = 1/2$.

Explanation:

1. Find the Principal Value ($\alpha$): We can rewrite this as $$\sin^2x = (\frac{1}{\sqrt{2}})^2 = \sin^2(\pi/4)$$ So, our reference angle is $\alpha = \pi/4$.

2. State the General Formula: The general solution for $\sin^2x = \sin^2\alpha$ (as well as for cos and tan) is $x = n\pi \pm \alpha$.

3. Substitute: $x = n\pi \pm \pi/4$.

Answer: $x = n\pi \pm \pi/4, n \in \mathbb{Z}$.

Example 7: Solving a Quadratic Equation

Question: Solve $2\cos^2x + 3\sin x = 0$.

Explanation:

1. Convert to a single function: Use the identity $\cos^2x = 1-\sin^2x$. The equation becomes $$2(1-\sin^2x) + 3\sin x = 0$$

2. Form a quadratic equation: $$2-2\sin^2x+3\sin x=0$$ $$\implies 2\sin^2x - 3\sin x - 2 = 0$$

3. Solve the quadratic: Let $y=\sin x$. We have $$2y^2-3y-2=0$$ which factors as $(2y+1)(y-2)=0$. The solutions are $y=-1/2$ and $y=2$.

4. Find valid solutions: This means $\sin x = -1/2$ or $\sin x = 2$.

   Since the range of sine is $[-1, 1]$, $\sin x = 2$ is impossible.

5. Find the general solution: For $\sin x = -1/2 = \sin(-\pi/6)$, the general solution is $$x = n\pi + (-1)^n(-\pi/6) = n\pi - (-1)^n(\pi/6)$$, which can also be written as $$x = n\pi + (-1)^{n+1}(\pi/6)$$

Answer: $x = n\pi + (-1)^{n+1} (\pi/6), n \in \mathbb{Z}$.

Example 8: Solving a Quadratic in Tangent

Question: Solve $\tan^2x + (1-\sqrt{3})\tan x - \sqrt{3} = 0$.

Explanation:

1. Form a quadratic: This is already a quadratic equation in terms of $\tan x$. Let $y=\tan x$, so we have $$y^2+(1-\sqrt{3})y-\sqrt{3}=0$$

2. Solve by factoring: By inspection or grouping, we can factor the equation: $$y^2+y-\sqrt{3}y-\sqrt{3}=0$$ $$\implies y(y+1)-\sqrt{3}(y+1)=0$$ $$\implies (y+1)(y-\sqrt{3})=0$$

3. Find trigonometric solutions: This gives $\tan x = -1$ or $\tan x = \sqrt{3}$.

4. Find general solutions:

   • For $\tan x = -1$, the principal value is $-\pi/4$. The general solution is $x=n\pi - \pi/4$.

   • For $\tan x = \sqrt{3}$, the principal value is $\pi/3$. The general solution is $x=n\pi + \pi/3$.

Answer: The solutions are $x=n\pi - \pi/4$ and $x=n\pi + \pi/3$.

Example 9: Using Sum-to-Product

Question: Solve $\sin x + \sin(3x) + \sin(5x) = 0$.

Explanation:

1. Group and apply sum-to-product: Group the outer terms $(\sin(5x)+\sin x)$. Use the identity $$\sin A + \sin B = 2\sin((A+B)/2)\cos((A-B)/2)$$ $$2\sin(3x)\cos(2x) + \sin(3x) = 0$$

2. Factor: Factor out the common term $\sin(3x)$.

   $\sin(3x)(2\cos(2x)+1)=0$.

3. Solve the two resulting equations:

   • Case 1: $\sin(3x)=0$. This implies $3x=n\pi$, so $x=n\pi/3$.
   • Case 2: $2\cos(2x)+1=0 \implies \cos(2x)=-1/2$.

   The principal value for $2x$ is $2\pi/3$.

   The general solution is $2x=2n\pi \pm 2\pi/3$, so $x=n\pi \pm \pi/3$.

Answer: The solutions are $x=n\pi/3$ and $x=n\pi \pm \pi/3$.

Example 10: Using Sum-to-Product

Question: Solve $\cos x + \cos(2x) + \cos(3x) = 0$.

Explanation:

1. Group and apply sum-to-product: Group $(\cos(3x)+\cos x)$. Use $$\cos A + \cos B = 2\cos((A+B)/2)\cos((A-B)/2)$$

   $$2\cos(2x)\cos x + \cos(2x) = 0$$

2. Factor: $\cos(2x)(2\cos x+1)=0$.

3. Solve the two resulting equations:

   • Case 1: $\cos(2x)=0$. This implies $2x=(2n+1)\pi/2$, so $x=(2n+1)\pi/4$.

   • Case 2: $2\cos x+1=0 \implies \cos x=-1/2$.

   The general solution is $x=2n\pi \pm 2\pi/3$.

Answer: The solutions are $x=(2n+1)\pi/4$ and $x=2n\pi \pm 2\pi/3$.

Example 11: Using Double Angle Identity

Question: Solve $\sin x \cos x = 1/4$.

Explanation:

1. Recognize the Double Angle Identity: The expression resembles the identity $$\sin(2x) = 2\sin x \cos x$$ Multiply the equation by 2. $$2\sin x \cos x = 2(1/4)$$ $$\implies \sin(2x)=1/2$$

2. Find the General Solution: We solve for the angle $2x$. The principal value for $\sin\theta=1/2$ is $\alpha=\pi/6$. The general solution is:

   $2x = n\pi + (-1)^n(\pi/6)$.

3. Solve for x: Divide the entire solution by 2.

Answer: $x = n\pi/2 + (-1)^n(\pi/12), n \in \mathbb{Z}$.

Example 12: Equation with tan and cot

Question: Solve $\tan x + \cot x = 2$.

Explanation:

1. Convert to Sine and Cosine: $$\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = 2$$

2. Simplify: Combine the fractions on the left side. $$\frac{\sin^2x + \cos^2x}{\sin x \cos x} = 2$$ $$\implies \frac{1}{\sin x \cos x} = 2$$

3. Use Double Angle Identity: Rearrange to get $$1=2\sin x \cos x$$ which is the identity for $\sin(2x)$. $\sin(2x) = 1$.

4. Find General Solution: The general solution for $\sin\theta=1$ is $\theta = 2n\pi + \pi/2$.

   $2x = 2n\pi + \pi/2$. Dividing by 2 gives the final answer.

Answer: $x = n\pi + \pi/4$.

Example 13: Homogeneous Equation

Question: Solve $3\sin^2x - 5\sin x\cos x - 2\cos^2x = 0$.

Explanation:

1. Identify as Homogeneous: An equation involving terms of the same degree in $\sin x$ and $\cos x$ is homogeneous. We can convert it to a quadratic in $\tan x$.

2. Divide by $\cos^2x$: Assuming $\cos x \ne 0$, we divide the entire equation by $\cos^2x$. $$3\frac{\sin^2x}{\cos^2x} - 5\frac{\sin x\cos x}{\cos^2x} - 2\frac{\cos^2x}{\cos^2x} = 0$$ $$\implies 3\tan^2x-5\tan x -2=0$$

3. Solve the Quadratic: Let $y=\tan x$. $3y^2-5y-2=0$. Factoring gives $$(3y+1)(y-2)=0$$ So, $y=-1/3$ or $y=2$.

4. Find General Solutions:

   • For $\tan x = 2$, the solution is $x=n\pi+\tan^{-1}(2)$.

   • For $\tan x = -1/3$, the solution is

   $x=n\pi+\tan^{-1}(-1/3)$ or $x=n\pi-\tan^{-1}(1/3)$.

Answer: $x=n\pi+\tan^{-1}(2)$ and $x=n\pi-\tan^{-1}(1/3)$.

Example 14: Using Identities

Question: Solve $\cos(2x) = \cos^2x$.

Explanation:

1. Use Double Angle Identity: Substitute the identity $\cos(2x)=2\cos^2x-1$.

   $2\cos^2x-1=\cos^2x$.

2. Solve for $\cos^2x$:

   $\cos^2x = 1$.

3. Solve for $\cos x$:

   This gives two possibilities: $\cos x = 1$ or $\cos x = -1$.

4. Find General Solutions:

   • $\cos x = 1$ has the solution $x=2n\pi$.

   • $\cos x = -1$ has the solution $x=(2n+1)\pi$.

   Together, these are all the integer multiples of $\pi$.

Answer: $x=n\pi, n \in \mathbb{Z}$.

Example 15: Solutions in an Interval

Question: Find the number of solutions of $\tan x + \sec x = 2\cos x$ in $[0, 2\pi]$.

Explanation:

1. Convert and Simplify: $$\frac{\sin x+1}{\cos x}=2\cos x$$

   This requires $\cos x \ne 0$.

   $$\sin x+1 = 2\cos^2x = 2(1-\sin^2x)$$ $$\implies 2\sin^2x + \sin x - 1 = 0$$

2. Solve the Quadratic: Factoring gives $$(2\sin x-1)(\sin x+1)=0$$

   This yields $\sin x = 1/2$ or $\sin x = -1$.

3. Find Solutions and Check Validity:

   • For $\sin x = 1/2$, the solutions in $[0, 2\pi]$ are $x = \pi/6$ and $x = 5\pi/6$. At these values, $\cos x \ne 0$, so they are valid.

   • For $\sin x = -1$, the solution is $x = 3\pi/2$. At this value, $\cos(3\pi/2) = 0$. This violates the condition from step 1, so it is an extraneous solution.

Answer: There are 2 solutions.

Example 16: Solving $a\sin x + b\cos x = c$

Question: Solve $\sin x + \cos x = 1$.

Explanation:

1. Normalize the Equation: Divide by $$\sqrt{a^2+b^2} = \sqrt{1^2+1^2} = \sqrt{2}$$

   $$\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x = \frac{1}{\sqrt{2}}$$

2. Convert to a Single Function: This can be seen as $$\cos(\pi/4)\sin x + \sin(\pi/4)\cos x = \sin(x+\pi/4)$$, or as $$\cos x\cos(\pi/4)+\sin x\sin(\pi/4) = \cos(x-\pi/4)$$ Let's use the cosine form. $$\cos(x-\pi/4) = 1/\sqrt{2} = \cos(\pi/4)$$

3. Find General Solution: Using the formula for $$\cos A = \cos B$$ $$\implies A=2n\pi \pm B$$

   $x-\pi/4 = 2n\pi \pm \pi/4$.

4. Separate the Cases:

   • $x-\pi/4 = 2n\pi + \pi/4 \implies x=2n\pi+\pi/2$.
   • $x-\pi/4 = 2n\pi - \pi/4 \implies x=2n\pi$.

Answer: $x=2n\pi+\pi/2$ and $x=2n\pi$.

Example 17: Solving $a\sin x + b\cos x = c$

Question: Solve $\sin x + \sqrt{3}\cos x = \sqrt{2}$.

Explanation:

1. Normalize: Divide by $$\sqrt{1^2+(\sqrt{3})^2} = 2$$ $$\frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x$$ $$= \frac{\sqrt{2}}{2}$$ $$= \frac{1}{\sqrt{2}}$$

2. Convert: This can be written as $$\cos(\pi/3)\sin x + \sin(\pi/3)\cos x = \sin(x+\pi/3)$$, or as $$\sin(\pi/6)\sin x + \cos(\pi/6)\cos x = \cos(x-\pi/6)$$ Let's use the cosine form. $$\cos(x-\pi/6) = 1/\sqrt{2} = \cos(\pi/4)$$

3. Solve: $x-\pi/6 = 2n\pi \pm \pi/4$.

   • $x = 2n\pi + \pi/4 + \pi/6 = 2n\pi + 5\pi/12$.

   • $x = 2n\pi - \pi/4 + \pi/6 = 2n\pi - \pi/12$.

Answer: $x = 2n\pi + 5\pi/12$ and $x = 2n\pi - \pi/12$.

Example 18: Solutions in an Interval

Question: Find the number of solutions of $\cos x + \sqrt{3}\sin x = 2$ in $[0, 2\pi]$.

Explanation:

1. Normalize: Divide by $\sqrt{1^2+(\sqrt{3})^2} = 2$.

   $\frac{1}{2}\cos x + \frac{\sqrt{3}}{2}\sin x = 1$.

2. Convert: This can be written as $$\sin(\pi/6)\cos x + \cos(\pi/6)\sin x = \sin(x+\pi/6)$$ So, $\sin(x+\pi/6)=1$.

3. Find General Solution: The general solution for $\sin\theta=1$ is $\theta = 2n\pi+\pi/2$.

   $x+\pi/6 = 2n\pi+\pi/2 \implies x=2n\pi+\pi/3$.

4. Find Solution in Interval: For $n=0$, we get $x=\pi/3$. For other integer values of $n$, $x$ is outside the interval $[0, 2\pi]$.

Answer: There is only 1 solution.

Example 19: Range of $a\cos x + b\sin x$

Question: For what values of k does the equation $3\cos x + 4\sin x = k$ have a solution?

Explanation:

An expression of the form $a\cos x + b\sin x$ can be converted to $R\cos(x-\alpha)$ where $R = \sqrt{a^2+b^2}$.

Here, $a=3, b=4$, so $R = \sqrt{3^2+4^2}=5$. The expression becomes $5\cos(x-\alpha)$.

The range of the cosine function is $[-1, 1]$. Therefore, the range of $5\cos(x-\alpha)$ is $[-5, 5]$.

For the equation $5\cos(x-\alpha)=k$ to have a solution, the value of $k$ must be within this range.

Answer: The solution exists for $k \in [-5, 5]$.

Example 20: Solving an Equation with sec and tan

Question: Solve $3\tan x + 4 = 5\sec x$.

Explanation:

1. Convert to sin/cos: $$3\frac{\sin x}{\cos x} + 4 = \frac{5}{\cos x}$$ Multiply by $\cos x$ (assuming $\cos x \ne 0$).

   $3\sin x + 4\cos x = 5$.

2. Normalize: Divide by $\sqrt{3^2+4^2}=5$.

   $\frac{3}{5}\sin x + \frac{4}{5}\cos x = 1$.

3. Convert to a Single Function: Let $\alpha = \cos^{-1}(3/5)$. Then $\sin\alpha = 4/5$. The equation is $$\sin x \cos\alpha + \cos x \sin\alpha = 1$$, which is $\sin(x+\alpha)=1$.

4. Solve: The general solution for $\sin\theta=1$ is $\theta = 2n\pi+\pi/2$.

   $x+\alpha = 2n\pi+\pi/2$.

Answer: $x=2n\pi+\pi/2-\alpha$, where $\alpha=\cos^{-1}(3/5)$.

Example 21: Solving $\cos A = \sin B$

Question: Find the number of solutions of $\cos(3x) = \sin(2x)$ in $[0, 2\pi]$.

Explanation:

1. Convert: Use $\sin(2x) = \cos(\pi/2-2x)$. The equation is $$\cos(3x) = \cos(\pi/2-2x)$$

2. Find General Solution: For $\cos A = \cos B$, $A=2n\pi \pm B$.

   • Case 1: $$3x = 2n\pi + (\pi/2-2x)$$ $$\implies 5x = (4n+1)\pi/2$$ $$\implies x = (4n+1)\pi/10$$

   • Case 2: $$3x = 2n\pi - (\pi/2-2x)$$ $$\implies x = 2n\pi - \pi/2$$

3. Find Solutions in the Interval:

   • From Case 1: For $n=0,1,2,3,4$, we get $$x=\pi/10, 5\pi/10=\pi/2, 9\pi/10, 13\pi/10, 17\pi/10 (5 solutions)$$
   • From Case 2: For $n=1$, we get $x=2\pi-\pi/2 = 3\pi/2$. (1 solution).

Answer: There are a total of 6 solutions.

Example 22: Using Extremal Values

Question: Solve $\cos x + \cos y - \cos(x+y) = 3/2$.

Explanation: The maximum value of the expression $$\cos x + \cos y - \cos(x+y)$$ is $3/2$, and this occurs only when $\cos x = 1/2$, $\cos y = 1/2$, and $\cos(x+y)=-1/2$.

A simple case is $x=\pi/3, y=\pi/3$.

Then $\cos(x+y)=\cos(2\pi/3)=-1/2$.

The equation becomes $1/2+1/2-(-1/2)=3/2$, which is true.

To find the general solution, we solve for when these conditions are met.

One family of solutions is $x=2n\pi+\pi/3$ and $y=2m\pi+\pi/3$ (and vice versa).

Answer: One set of solutions is $x=2n\pi+\pi/3, y=2m\pi+\pi/3$ for integers n,m.

Example 23: Graphical Solution

Question: Find the number of solutions for $\sin x = x/10$.

Explanation: We solve this by finding the number of intersection points of the graphs $y=\sin x$ and $y=x/10$.

1. Analyze the graphs: $y=\sin x$ is a wave oscillating between -1 and 1. $y=x/10$ is a straight line passing through the origin with a slope of 0.1.

2. Bound the search: Since $-1 \le \sin x \le 1$, we only need to look for intersections where $-1 \le x/10 \le 1$, which means $-10 \le x \le 10$.

3. Sketch and Count:

   • The graphs will intersect at the origin ($x=0$).
   • For $x>0$: The line $y=x/10$ will intersect the sine curve. Since $10 \approx 3.18\pi$, we check the sine curve up to $3.18\pi$. The sine curve completes one full wave at $2\pi \approx 6.28$ and another half-wave at $3\pi \approx 9.42$. The line will intersect this curve twice in $(0, 2\pi)$ and once more in $(2\pi, 3\pi)$. Total 3 positive solutions.
   • For $x<0$: Due to the symmetry of both functions, there will also be 3 negative solutions.

Total solutions = 3 (positive) + 3 (negative) + 1 (at origin) = 7.

Answer: 7 solutions.

Example 24: Solving with Domain Check

Question: Solve $\sin x = \tan x$.

Explanation:

1. Rewrite and Factor:

   $\sin x = \sin x / \cos x$

   $\implies \sin x - \sin x / \cos x = 0$

   $\implies \sin x (1-1/\cos x)=0$.

2. Find Potential Solutions:

   • Case 1:

   $\sin x = 0 \implies x=n\pi$.

   • Case 2:

   $1-1/\cos x = 0 \implies \cos x = 1 \implies x=2n\pi$.

3. Combine Solutions: The solution set from Case 2 ($x=2n\pi$) is a subset of the solutions from Case 1 ($x=n\pi$). Therefore, the complete general solution is given by Case 1.

4. Check Domain: The original equation has $\tan x$, which is undefined when $\cos x = 0$. Our solution $x=n\pi$ always has $\cos x = \pm 1$, so it is valid.

Answer: $x=n\pi, n \in \mathbb{Z}$.

Example 25: Sum of Solutions in an Interval

Question: Find the sum of solutions of $\cos(2x)=\sin x$ in $[0,2\pi]$.

Explanation:

1. Form a Quadratic: Use the identity $\cos(2x)=1-2\sin^2x$.

   The equation becomes $$1-2\sin^2x=\sin x$$ $$\implies 2\sin^2x+\sin x-1=0$$

2. Solve for $\sin x$: Factoring gives $$(2\sin x-1)(\sin x+1)=0$$ So, $\sin x=1/2$ or $\sin x=-1$.

3. Find Solutions in $[0,2\pi]$:

   • For $\sin x=1/2$, the solutions are $x=\pi/6$ and $x=5\pi/6$.
   • For $\sin x=-1$, the solution is $x=3\pi/2$.

4. Calculate the Sum:

   Sum = $\pi/6 + 5\pi/6 + 3\pi/2 = 6\pi/6 + 3\pi/2 = \pi+3\pi/2 = 5\pi/2$.

Answer: $5\pi/2$.

Example 26: Using Double Angle Identity

Question: Solve $\sin^4x + \cos^4x = 1/2$.

Explanation:

1. Use Algebraic Identity: $$\sin^4x + \cos^4x = (\sin^2x+\cos^2x)^2 - 2\sin^2x\cos^2x$$ $$= 1 - 2\sin^2x\cos^2x$$

2. Simplify: The equation becomes $$1-2\sin^2x\cos^2x = 1/2$$ $$\implies 2\sin^2x\cos^2x=1/2$$ $$\implies 4\sin^2x\cos^2x=1$$ $$\implies (2\sin x\cos x)^2=1$$ $$\implies \sin^2(2x)=1$$

3. Solve: This means $\sin(2x)=\pm 1$. This occurs when $2x$ is an odd multiple of $\pi/2$.

   So, $2x = (2n+1)\pi/2$.

Answer: $x=(2n+1)\pi/4$.

Example 27: Solving a Complex Equation

Question: Solve the equation $(1-\tan\theta)(1+\sin 2\theta) = 1+\tan\theta$.

Explanation:

1. Use Identities: Rewrite $$1+\sin 2\theta = 1+2\sin\theta\cos\theta$$ $$= \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta$$ $$= (\sin\theta+\cos\theta)^2$$ Also rewrite tan in terms of sin and cos. $$(1-\frac{\sin\theta}{\cos\theta})(\sin\theta+\cos\theta)^2 = 1+\frac{\sin\theta}{\cos\theta}$$ $$\frac{\cos\theta-\sin\theta}{\cos\theta}(\sin\theta+\cos\theta)^2 = \frac{\cos\theta+\sin\theta}{\cos\theta}$$
2. Solve: Assuming $\cos\theta \ne 0$, we can cancel the denominator. $$(\cos\theta-\sin\theta)(\sin\theta+\cos\theta)^2 - (\cos\theta+\sin\theta) = 0$$ Factor out $(\cos\theta+\sin\theta)$: $$(\cos\theta+\sin\theta)[(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)-1]=0$$ $$(\cos\theta+\sin\theta)[\cos^2\theta-\sin^2\theta-1]=0$$ $$(\cos\theta+\sin\theta)[\cos(2\theta)-1]=0$$
3. Find General Solutions:
   • Case 1: $\cos\theta+\sin\theta=0 \implies \tan\theta=-1 \implies \theta = n\pi - \pi/4$.
   • Case 2: $\cos(2\theta)-1=0 \implies \cos(2\theta)=1 \implies 2\theta = 2n\pi \implies \theta = n\pi$.

Answer: $x=n\pi - \pi/4$ and $x=n\pi$.

Example 28: Simple General Solution

Question: Find the general solution of $\sin^2(2x)=0$.

Explanation:

1. Simplify: If the square of a number is zero, the number itself must be zero. So, the equation simplifies to $\sin(2x)=0$.

2. Apply General Formula: The general solution for $\sin\theta=0$ is $\theta = n\pi$.

3. Solve for x: Here, $\theta=2x$. So, $2x = n\pi \implies x = n\pi/2$.

Answer: $x = n\pi/2, n \in \mathbb{Z}$.

Example 29: Using Range of Sine

Question: Solve the equation $\sin x + \sin y = 2$.

Explanation: The maximum value of the sine function is 1. The only way for the sum of two sine terms to be 2 is if both terms are at their maximum value simultaneously.

• $\sin x = 1 \implies x = 2n\pi + \pi/2$.
• $\sin y = 1 \implies y = 2m\pi + \pi/2$.

(We use different integers $n$ and $m$ as $x$ and $y$ are independent variables).

Answer: $x=2n\pi+\pi/2$ and $y=2m\pi+\pi/2$ for integers n, m.

Example 30: Using Range and GIF

Question: Find the number of solutions of $|\cos x| = 2[x]$ where [.] is the Greatest Integer Function.

Explanation:

1. Analyze the LHS: The range of $|\cos x|$ is $[0,1]$.

2. Analyze the RHS: The Greatest Integer Function, $[x]$, produces only integer values. Therefore, $2[x]$ can only produce even integer values (…, -4, -2, 0, 2, 4, …).

3. Find Common Ground: We need to find a value that is present in both the range of the LHS and the possible values of the RHS. The only value that satisfies both conditions is 0.

4. Solve the System of Equations: We must solve for $x$ where both sides are simultaneously zero.

   • Eq 1: $|\cos x|=0 \implies \cos x = 0$. This occurs at $x = (2n+1)\pi/2$.
   • Eq 2: $2[x]=0 \implies [x]=0$. This is true for all $x$ in the interval $0 \le x < 1$.

5. Find Intersection: We need to find if any solution from Eq 1 falls within the interval from Eq 2. The smallest positive solution for $\cos x=0$ is $x=\pi/2 \approx 1.57$. This is not in the interval $[0,1)$. There are no other solutions in this interval.

Answer: There are no solutions.