Question and Answer Section

Example 1: Finding Principal Solutions (Sine)

Question: Find the principal solutions of the equation $\sin x = 1/2$.

Solution: Principal solutions are the values of x in the interval $[0, 2\pi)$.

Step 1: Identify the quadrants. Sine is positive in the first (Q1) and second (Q2) quadrants.

Step 2: Find the angle in Q1. The base angle for which $\sin x = 1/2$ is $x = \pi/6$.

Step 3: Find the angle in Q2. The angle is $\pi - (\text{base angle}) = \pi - \pi/6 = 5\pi/6$.

The principal solutions are $\pi/6$ and $5\pi/6$.

Example 2: Finding Principal Solutions (Tangent)

Question: Find the principal solutions of the equation $\tan x = -1/\sqrt{3}$.

Solution: Step 1: Identify the quadrants. Tangent is negative in the second (Q2) and fourth (Q4) quadrants.

Step 2: Find the reference angle $\alpha$ for which $\tan \alpha = 1/\sqrt{3}$. This is $\alpha = \pi/6$.

Step 3: Find the angle in Q2: $x = \pi - \alpha = \pi - \pi/6 = 5\pi/6$.

Step 4: Find the angle in Q4: $x = 2\pi - \alpha = 2\pi - \pi/6 = 11\pi/6$.

The principal solutions are $5\pi/6$ and $11\pi/6$.

Example 3: Finding General Solution (Cosine)

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

Solution: Step 1: Find the smallest positive angle (principal value) $\alpha$ that satisfies the equation. Cosine is negative in Q2 and Q3. The reference angle for $\cos \theta = 1/2$ is $\pi/3$. The smallest positive angle is in Q2, so $\alpha = \pi - \pi/3 = 2\pi/3$.

Step 2: The equation is now $\cos x = \cos(2\pi/3)$.

Step 3: Apply the general solution formula for cosine, $x = 2n\pi \pm \alpha$. The solution is $x = 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z}$.

Example 4: Finding General Solution (Sine)

Question: Find the general solution of $\sin(2x) = -\sqrt{3}/2$.

Solution: Step 1: Find the principal value $\alpha$. Sine is negative in Q3 and Q4. The reference angle is $\pi/3$. The smallest positive angle is in Q3, so $\alpha = \pi + \pi/3 = 4\pi/3$.

Step 2: The equation becomes $\sin(2x) = \sin(4\pi/3)$.

Step 3: Apply the general solution for sine, $A = n\pi + (-1)^n B$. $2x = n\pi + (-1)^n (4\pi/3)$.

Step 4: Solve for x. $x = \frac{n\pi}{2} + (-1)^n \frac{2\pi}{3}, n \in \mathbb{Z}$.

Example 5: Finding General Solution (Tangent)

Question: Solve the equation $\tan(3x) = 1$.

Solution: Step 1: Find the principal value $\alpha$. We know $\tan(\pi/4)=1$. So $\alpha=\pi/4$.

Step 2: The equation becomes $\tan(3x) = \tan(\pi/4)$.

Step 3: Apply the general solution for tangent, $A = n\pi + B$. $3x = n\pi + \pi/4$.

Step 4: Solve for x. $x = \frac{n\pi}{3} + \frac{\pi}{12}, n \in \mathbb{Z}$.

Example 6: Solving with Squared Functions

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

Solution: Step 1: Find the principal angle $\alpha$. We have $\cos^2x = (1/2)^2$. We know $\cos(\pi/3) = 1/2$, so we can write the equation as $\cos^2x = \cos^2(\pi/3)$.

Step 2: Use the general solution formula for squared functions, $x=n\pi \pm \alpha$. So, $x = n\pi \pm \frac{\pi}{3}, n \in \mathbb{Z}$.

Example 7: Quadratic in $\sin x$

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

Solution: Step 1: Use $\sin^2x = 1-\cos^2x$ to get a quadratic in $\cos x$. $2(1-\cos^2x) + 3\cos x = 0 \implies 2\cos^2x-3\cos x-2=0$.

Step 2: Let $y=\cos x$. The equation is $2y^2-3y-2=0$, which factors to $(2y+1)(y-2)=0$.

Step 3: The solutions are $\cos x = -1/2$ or $\cos x = 2$. Since the range of cosine is $[-1,1]$, $\cos x = 2$ is impossible.

Step 4: Solve $\cos x = -1/2$. The principal value is $2\pi/3$. The general solution is $x = 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z}$.

Example 8: Quadratic in $\tan x$

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

Solution: This is a quadratic equation in $\tan x$. Let $y = \tan x$: $y^2 - (1+\sqrt{3})y + \sqrt{3} = 0$. This can be factored as $(y-1)(y-\sqrt{3})=0$.

• Case 1: $\tan x = 1 = \tan(\pi/4)$. General solution is $x = n\pi + \pi/4$.
• Case 2: $\tan x = \sqrt{3} = \tan(\pi/3)$. General solution is $x = n\pi + \pi/3$.

Example 9: Solving $a\cos x + b\sin x = c$

Question: Find the general solution of $\sqrt{3}\cos x + \sin x = \sqrt{2}$.

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

Step 2: Convert the LHS to a single cosine term using $\cos(A-B) = \cos A \cos B + \sin A \sin B$. $\cos(\pi/6)\cos x + \sin(\pi/6)\sin x = \cos(x-\pi/6)$.

Step 3: Solve the resulting equation. $\cos(x-\pi/6) = \frac{1}{\sqrt{2}} = \cos(\pi/4)$. The general solution is $x-\pi/6 = 2n\pi \pm \pi/4$. This gives two sets of solutions:

• $x = 2n\pi + \pi/4 + \pi/6 = \mathbf{2n\pi + 5\pi/12}$.
• $x = 2n\pi - \pi/4 + \pi/6 = \mathbf{2n\pi - \pi/12}$.

Example 10: Using Sum-to-Product

Question: Solve $\sin(2x) + \sin(4x) + \sin(6x) = 0$.

Solution: Step 1: Group the outer terms: $(\sin(6x)+\sin(2x)) + \sin(4x) = 0$.

Step 2: Apply the sum-to-product formula: $2\sin(\frac{6x+2x}{2})\cos(\frac{6x-2x}{2}) + \sin(4x) = 0 \implies 2\sin(4x)\cos(2x) + \sin(4x) = 0$.

Step 3: Factor and solve. $\sin(4x)(2\cos(2x)+1)=0$. This gives two possibilities:

• $\sin(4x)=0 \implies 4x=n\pi \implies \mathbf{x = n\pi/4}$.
• $2\cos(2x)+1=0 \implies \cos(2x)=-1/2 = \cos(2\pi/3)$. So, $2x = 2n\pi \pm 2\pi/3 \implies \mathbf{x=n\pi \pm \pi/3}$.

Example 11: Solving by Squaring

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

Solution: Step 1: Square both sides: $(\sin x + \cos x)^2 = 1^2 \implies \sin^2x+\cos^2x+2\sin x\cos x = 1$.

Step 2: Simplify: $1 + \sin(2x) = 1 \implies \sin(2x) = 0$.

Step 3: Find the general solution of the new equation: $2x = n\pi$, so potential solutions are $x = n\pi/2$.

Step 4: Check for extraneous roots by testing the solutions in the original equation.

• $x=0$: $\sin 0 + \cos 0 = 0+1=1$ (Correct).
• $x=\pi/2$: $\sin(\pi/2)+\cos(\pi/2)=1+0=1$ (Correct).
• $x=\pi$: $\sin(\pi)+\cos(\pi)=0+(-1)=-1$ (False).
• $x=3\pi/2$: $\sin(3\pi/2)+\cos(3\pi/2)=-1+0=-1$ (False). The valid solutions repeat every $2\pi$. They are of the form $x=2n\pi$ and $x=2n\pi+\pi/2$.

Example 12: Equations with $\tan$ and $\cot$

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

Solution: Step 1: Rewrite in terms of sine and cosine: $\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = 2$.

Step 2: Find a common denominator: $\frac{\sin^2x+\cos^2x}{\sin x \cos x} = 2 \implies \frac{1}{(1/2)\sin(2x)} = 2$.

Step 3: Solve for x. $\frac{2}{\sin(2x)}=2 \implies \sin(2x) = 1$. The general solution for $2x$ is $2x = 2n\pi + \pi/2$. $x = \mathbf{n\pi + \pi/4, n \in \mathbb{Z}}$.

Example 13: Homogeneous Equations

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

Solution: Step 1: This is a homogeneous equation. Assuming $\cos x \ne 0$, we can divide the entire equation by $\cos^2x$. $$2\frac{\sin^2x}{\cos^2x} - 5\frac{\sin x \cos x}{\cos^2x} - 3\frac{\cos^2x}{\cos^2x} = 0$$ $$2\tan^2x - 5\tan x - 3 = 0$$

Step 2: Let $y=\tan x$ and solve the quadratic: $2y^2-5y-3=0 \implies (2y+1)(y-3)=0$.

Step 3: The solutions are $\tan x = -1/2$ or $\tan x = 3$. The general solution is $x=n\pi+\tan^{-1}(-1/2)$ or $x=n\pi+\tan^{-1}(3)$, where $n \in \mathbb{Z}$.

Example 14: Using Multiple Angle Formulas

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

Solution: Step 1: Use the sum-to-product formula on the first two terms: $2\cos(\frac{3x+x}{2})\cos(\frac{3x-x}{2}) - 2\cos(2x) = 0$. $2\cos(2x)\cos(x) - 2\cos(2x) = 0$.

Step 2: Factor out the common term $2\cos(2x)$: $2\cos(2x)(\cos x - 1) = 0$.

Step 3: This gives two possibilities:

• $\cos(2x) = 0 \implies 2x = (2n+1)\pi/2 \implies \mathbf{x=(2n+1)\pi/4}$.
• $\cos x - 1 = 0 \implies \cos x = 1 \implies \mathbf{x = 2n\pi}$.

Example 15: Finding Number of Solutions in an Interval

Question: Find the number of solutions of the equation $2\sin^2x + 5\sin x - 3 = 0$ in the interval $[0, 3\pi]$.

Solution: Step 1: Let $y=\sin x$. The equation is $2y^2+5y-3=0 \implies (2y-1)(y+3)=0$. Step 2: The solutions are $y=1/2$ or $y=-3$. Since $-1 \le \sin x \le 1$, the only valid case is $\sin x = 1/2$. Step 3: Find the solutions for $\sin x = 1/2$ in the interval $[0, 3\pi]$.

• In $[0, 2\pi]$, the solutions are $\mathbf{\pi/6}$ and $\mathbf{5\pi/6}$.
• In $(2\pi, 3\pi]$, the solutions are $2\pi+\pi/6 = \mathbf{13\pi/6}$ and $2\pi+5\pi/6 = \mathbf{17\pi/6}$. There are a total of 4 solutions.

Example 16: Solving an Equation with sec and tan

Question: Solve $\sec^2(2x) = 1 - \tan(2x)$.

Solution: Step 1: Use the Pythagorean identity $\sec^2\theta = 1+\tan^2\theta$. The equation becomes $1+\tan^2(2x) = 1-\tan(2x)$.

Step 2: Simplify and solve the resulting equation. $\tan^2(2x)+\tan(2x)=0$. Factor out $\tan(2x)$: $\tan(2x)(\tan(2x)+1)=0$.

Step 3: This gives two cases:

• $\tan(2x)=0 \implies 2x=n\pi \implies \mathbf{x=n\pi/2}$.
• $\tan(2x)=-1 \implies 2x=n\pi- \pi/4 \implies \mathbf{x=n\pi/2-\pi/8}$.

Example 17: Solving $\sin A = \sin B$ type

Question: Find the general solution for $\sin(3x) = \sin(x)$.

Solution: Using the general solution $A = n\pi + (-1)^n B$, we get $3x = n\pi + (-1)^n x$. We must solve for even and odd n separately.

• Case 1: n is even (n=2k). $3x=2k\pi+(-1)^{2k}x = 2k\pi+x$. $2x=2k\pi \implies \mathbf{x=k\pi}$.

• Case 2: n is odd (n=2k+1). $3x=(2k+1)\pi+(-1)^{2k+1}x = (2k+1)\pi-x$. $4x=(2k+1)\pi \implies \mathbf{x=(2k+1)\pi/4}$. The complete set of solutions is $x=n\pi$ and $x=(2n+1)\pi/4$ for any integer n.

Example 18: Solving with Maximum/Minimum Values

Question: Find the number of solutions for the equation $\sin(2x) + \cos(4x) = 2$ in $[0, 2\pi]$.

Solution: The maximum value of $\sin(2x)$ is 1 and of $\cos(4x)$ is 1. For their sum to be 2, both must be 1 simultaneously.

• We need $\sin(2x)=1$, which implies $2x = 2n\pi+\pi/2$.
• We also need $\cos(4x)=1$, which implies $4x=2k\pi \implies 2x=k\pi$. For a solution to exist, we would need $k\pi = 2n\pi+\pi/2$, which implies $(k-2n)=1/2$. This is impossible for integers k and n. There are no solutions.

Example 19: Checking Domain

Question: Find the number of solutions for $\frac{\tan 3x - \tan 2x}{1+\tan 3x \tan 2x} = 1$ in $[0, 2\pi]$.

Solution: Step 1: The LHS is the formula for $\tan(3x-2x)=\tan x$. So we solve the simple equation $\tan x=1$. The solutions in $[0,2\pi]$ are $x=\pi/4$ and $x=5\pi/4$.

Step 2: We must check if these solutions are valid in the original equation. The original expression requires $\tan(2x)$ to be defined, which means $2x \neq (2n+1)\pi/2$.

• For $x=\pi/4$, $2x=\pi/2$. At this value, $\tan(2x)$ is undefined. So $x=\pi/4$ is an extraneous solution.
• For $x=5\pi/4$, $2x=5\pi/2$. At this value, $\tan(2x)$ is also undefined. So $x=5\pi/4$ is also extraneous. Therefore, there are no solutions.

Example 20: Using $\cos(2x)$ formulas

Question: Solve $3\cos(2x) + 7\cos x + 3 = 0$.

Solution: Step 1: Use the identity $\cos(2x) = 2\cos^2x-1$ to get a quadratic in $\cos x$. $3(2\cos^2x-1) + 7\cos x + 3 = 0 \implies 6\cos^2x-3 + 7\cos x + 3 = 0 \implies 6\cos^2x+7\cos x=0$.

Step 2: Factor and solve for $\cos x$. $\cos x(6\cos x + 7) = 0$. This gives two possibilities:

• $\cos x = 0$.
• $\cos x = -7/6$. This is impossible as it is outside the range [-1,1].

Step 3: Find the general solution for $\cos x = 0$. The general solution is $x = (2n+1)\pi/2, n \in \mathbb{Z}$.