Principal and General Solutions

1. Introduction to Trigonometric Equations

Just as an algebraic equation like $x^2 - 4 = 0$ asks, "What number, when squared, gives 4?", a trigonometric equation asks for an angle.

An equation involving one or more trigonometric functions of an unknown angle is called a trigonometric equation. For example, $\sin x = 1/2$ asks, "What angle $x$, when you take its sine, gives 1/2?".

Unlike algebraic equations which usually have a finite number of solutions, trigonometric equations have infinite solutions due to the periodic nature of sine, cosine, and tangent functions. We categorize these solutions into two types: principal and general. 🎯

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2. Principal Solutions

The solutions of a trigonometric equation that lie within one full rotation, i.e., in the interval $[0, 2\pi)$, are called principal solutions.

To find them, we use the ASTC rule (All Students Take Calculus) to determine the quadrants where the function has the required sign.

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

1. Identify Quadrants: $\sin x$ is positive, so solutions lie in Quadrant I and Quadrant II.
2. Find Base Angle ($\alpha$): The base angle in Q1 where $\sin x = 1/2$ is $x = \pi/6$.
3. Find Q2 Solution: The rule for Q2 is $\pi - \alpha$. So, the second solution is $x = \pi - \pi/6 = 5\pi/6$.
4. Result: The principal solutions are $\{\pi/6, 5\pi/6\}$.

Example 2: Find the principal solutions of $\cos x = -1/2$.

1. Identify Quadrants: $\cos x$ is negative, so solutions lie in Q2 and Q3.
2. Find Base Angle ($\alpha$): First, find the angle for the positive value: $\cos \alpha = 1/2 \implies \alpha = \pi/3$. This is our reference angle.
3. Find Q2 Solution: The rule is $\pi - \alpha$. So, $x = \pi - \pi/3 = 2\pi/3$.
4. Find Q3 Solution: The rule is $\pi + \alpha$. So, $x = \pi + \pi/3 = 4\pi/3$.
5. Result: The principal solutions are $\{2\pi/3, 4\pi/3\}$.

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3. General Solutions

The general solution is a formula that uses an integer '$n$' to represent all possible solutions to a trigonometric equation, including those outside the $[0, 2\pi)$ interval. The integer $n$ represents the number of full rotations made. 🔄

Formulas for General Solutions:

• If $\sin x = \sin \alpha$: The solutions are in Q1 ($...-2\pi+\alpha, \alpha, 2\pi+\alpha, ...$) and Q2 ($...-2\pi+(\pi-\alpha), \pi-\alpha, 2\pi+(\pi-\alpha), ...$). These two series are cleverly combined into a single formula: $$x = n\pi + (-1)^n \alpha, \quad n \in \mathbb{Z}$$ Example: For $\sin x = 1/2$, $\alpha = \pi/6$. The general solution is $x = n\pi + (-1)^n \frac{\pi}{6}$.

• If $\cos x = \cos \alpha$: The solutions are in Q1 ($"\alpha"$) and Q4 ($-"\alpha"$ or $2\pi - \alpha$). All other solutions are found by adding full rotations ($2\pi$) to these. $$x = 2n\pi \pm \alpha, \quad n \in \mathbb{Z}$$ Example: For $\cos x = 1/2$, $\alpha = \pi/3$. The general solution is $x = 2n\pi \pm \frac{\pi}{3}$.

• If $\tan x = \tan \alpha$: The solutions for tangent repeat every $\pi$ radians (not $2\pi$). The solutions are in Q1 ($"\alpha"$) and Q3 ($"\pi+\alpha"$). We can get all solutions by starting at $\alpha$ and adding integer multiples of $\pi$. $$x = n\pi + \alpha, \quad n \in \mathbb{Z}$$ Example: For $\tan x = 1$, $\alpha = \pi/4$. The general solution is $x = n\pi + \frac{\pi}{4}$.

General Solutions for Squared Functions:

If $\sin^2x = \sin^2\alpha$, $\cos^2x = \cos^2\alpha$, or $\tan^2x = \tan^2\alpha$, it implies that $\sin x = \pm \sin \alpha$, etc. This covers solutions in all four quadrants. All three equations share the same compact general solution: $$x = n\pi \pm \alpha, \quad n \in \mathbb{Z}$$ Example: Solve $\sin^2x = 1/4$. This is $\sin^2x = (1/2)^2 = \sin^2(\pi/6)$. Here, $\alpha = \pi/6$. The general solution is $x = n\pi \pm \frac{\pi}{6}$.

Example 1: Finding Principal Solutions (Sine)

Question: Find the principal solutions of the equation $\sin x = \frac{\sqrt{3}}{2}$.

Explanation: Principal solutions are the solutions that lie in the interval $[0, 2\pi)$.

1. Identify Quadrants: Since $\sin x$ is positive, the angle $x$ must be in Quadrant I or Quadrant II.

2. Find Quadrant I Solution: The base angle $\alpha$ for which $\sin \alpha = \frac{\sqrt{3}}{2}$ is $\alpha = \pi/3$. This is our first principal solution.

3. Find Quadrant II Solution: The formula for the angle in the second quadrant is $\pi - \alpha$. So, the second solution is $$x = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$

Answer: The principal solutions are $\pi/3$ and $2\pi/3$.

Example 2: Finding Principal Solutions (Tangent)

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

Explanation:

1. Identify Quadrants: Since $\tan x$ is negative, the angle $x$ must be in Quadrant II or Quadrant IV.

2. Find Reference Angle ($\alpha$): First, we find the base angle for the positive value: $\tan \alpha = 1$. This gives the reference angle $\alpha = \pi/4$.

3. Find Quadrant II Solution: The formula for the angle in the second quadrant is $\pi - \alpha$. So, $x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

4. Find Quadrant IV Solution: The formula for the angle in the fourth quadrant is $2\pi - \alpha$. So, $$x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$$

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

Example 3: Finding General Solution (Cosine)

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

Explanation:

1. Find a Principal Value ($\alpha$): First, find one solution. $\cos x$ is negative in Q2. The reference angle for $\cos = 1/2$ is $\pi/3$. The Q2 angle is $\alpha = \pi - \pi/3 = 2\pi/3$.

2. State the General Formula: The general solution for $\cos x = \cos \alpha$ is given by $x = 2n\pi \pm \alpha$, where $n$ is any integer.

3. Substitute and Finalize: Substitute the principal value $\alpha = 2\pi/3$ into the formula.

Answer: The general solution is $x = 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z}$.

Example 4: Finding General Solution (Sine)

Question: Solve the equation $\sin(2x) = -\sqrt{3}/2$.

Explanation:

1. Find a Principal Value ($\alpha$): We need an angle $\alpha$ such that $\sin \alpha = -\sqrt{3}/2$. Sine is negative in Q3 and Q4. A convenient choice is the negative angle in Q4, $\alpha = -\pi/3$.

2. Set up the General Solution: The general solution for $\sin \theta = \sin \alpha$ is $\theta = n\pi + (-1)^n \alpha$. Here, $\theta = 2x$.

   $2x = n\pi + (-1)^n (-\pi/3)$.

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

Answer: $$x = \frac{n\pi}{2} - (-1)^n \frac{\pi}{6}, n \in \mathbb{Z}$$

Example 5: Finding General Solution (Tangent)

Question: Find the general solution of $\tan(2x) = -\cot(x+\pi/3)$.

Explanation:

1. Transform the Equation: To solve, both sides must be the same trigonometric function. We use the identity $$\tan(\pi/2+\theta) = -\cot(\theta)$$

   $$\tan(2x) = \tan(\pi/2 + (x+\pi/3)) = \tan(x + 5\pi/6)$$

2. Apply the General Formula: The general solution for $\tan A = \tan B$ is $$A = n\pi + B$$

   $2x = n\pi + (x + 5\pi/6)$.

3. Solve for x:

   $2x - x = n\pi + 5\pi/6$.

Answer: $x = n\pi + 5\pi/6, n \in \mathbb{Z}$.

Example 6: Solving a Quadratic in $\sin x$

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

Explanation:

1. Convert to a Single Function: Use the Pythagorean identity $$\sin^2x = 1-\cos^2x$$

   $$2(1-\cos^2x) + 3\cos x = 0$$ $$\implies 2-2\cos^2x+3\cos x=0$$ $$\implies 2\cos^2x-3\cos x-2=0$$

2. Solve the Quadratic: Let $y=\cos x$.

The equation becomes $2y^2-3y-2=0$. Factoring this gives $$(2y+1)(y-2)=0$$

3. Find Valid Solutions: The solutions are $y=-1/2$ and $y=2$. This means $\cos x = -1/2$ or $\cos x = 2$.

   Since the range of cosine is $[-1, 1]$, the solution $\cos x=2$ is impossible.

4. Find the General Solution: We solve $\cos x = -1/2$. The principal value is $\alpha=2\pi/3$. The general solution for cosine is $x = 2n\pi \pm \alpha$.

Answer: $x = 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z}$.

Example 7: Solving with Squared Functions

Question: Solve $\tan^2x = 3$.

Explanation:

1. Find the Principal Value ($\alpha$): We can rewrite the equation as $\tan^2x = (\sqrt{3})^2$. We know that $\tan(\pi/3) = \sqrt{3}$. So, the equation is in the form $\tan^2x = \tan^2(\pi/3)$. Here, $\alpha=\pi/3$.

2. Apply the General Formula: The general solution for any of the squared forms ($"\sin^2x=\sin^2\alpha"$, etc.) is $x=n\pi \pm \alpha$.

3. State the Solution: Substituting $\alpha=\pi/3$ gives the final answer.

Answer: $x = n\pi \pm \frac{\pi}{3}, n \in \mathbb{Z}$.

Example 8: Equation with sec and tan

Question: Solve $\sec x - \tan x = \sqrt{3}$.

Explanation:

1. Convert to Sine and Cosine: $$\frac{1}{\cos x} - \frac{\sin x}{\cos x} = \sqrt{3}$$ $$\implies \frac{1-\sin x}{\cos x} = \sqrt{3}$$

2. Rearrange: $$1-\sin x = \sqrt{3}\cos x$$ $$\implies \sqrt{3}\cos x + \sin x = 1$$

   This is in the form $$a\cos x + b\sin x = c$$

3. Normalize the Equation:

   Divide by $\sqrt{a^2+b^2} = \sqrt{(\sqrt{3})^2+1^2} = \sqrt{4} = 2$.

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

4. Convert to a Single Function: We can use the sum formula for sine, $$\sin(A+B) = \sin A \cos B + \cos A \sin B$$ Let $A=x$ and $\cos B = 1/2$, $\sin B = \sqrt{3}/2$, which means $B=\pi/3$.

   $\sin(x+\pi/3) = 1/2$.

5. Find the General Solution: We know $1/2 = \sin(\pi/6)$.

So we solve $\sin(x+\pi/3) = \sin(\pi/6)$.

The general solution is $x+\pi/3 = n\pi + (-1)^n (\pi/6)$.

Answer: $x = n\pi - \pi/3 + (-1)^n (\pi/6), n \in \mathbb{Z}$.

Example 9: Solving $\sin A = \sin B$

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

Explanation:

1. Apply the General Formula: For an equation of the form $\sin A = \sin B$, the general solution is $$A = n\pi + (-1)^n B$$

   Here, $A=3x$ and $B=x$.

   $3x = n\pi + (-1)^n x$.

2. Solve for the two cases of n:

   • Case 1: n is even. Let $n=2k$ for some integer $k$. Then $(-1)^n = 1$.

     $3x=2k\pi+x \implies 2x=2k\pi \implies x=k\pi$.

   • Case 2: n is odd. Let $n=2k+1$ for some integer $k$. Then $(-1)^n = -1$.

     $3x=(2k+1)\pi-x \implies 4x=(2k+1)\pi \implies x=(2k+1)\pi/4$.

3. State the Combined Solution: We can express the two families of solutions using $n$ as the integer variable for convention.

Answer: The solutions are $x=n\pi$ and $x=(2n+1)\pi/4$, where $n \in \mathbb{Z}$.

Example 10: Finding Solutions in an Interval

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

Explanation:

1. Solve the Quadratic: Let $y=\sin x$.

   The equation becomes $2y^2+5y-3=0$.

   Factoring this gives $(2y-1)(y+3)=0$. The solutions are $y=1/2$ and $y=-3$.

   This means $\sin x = 1/2$ or $\sin x = -3$.

   Since the range of $\sin x$ is $[-1,1]$, the solution $\sin x = -3$ is impossible.

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

   The general solution is $x = n\pi + (-1)^n (\pi/6)$.

3. Find Specific Solutions by substituting values for n:

   • For $n=0: x = 0\pi + (-1)^0(\pi/6) = \pi/6$. (In the interval)
   • For $n=1: x = 1\pi + (-1)^1(\pi/6) = \pi - \pi/6 = 5\pi/6$. (In the interval)
   • For $n=2: x = 2\pi + (-1)^2(\pi/6) = 2\pi + \pi/6 = 13\pi/6$. (In the interval, since $3\pi = 18\pi/6$)
   • For $n=3: x = 3\pi + (-1)^3(\pi/6) = 3\pi - \pi/6 = 17\pi/6$. (In the interval)
   • For $n=4: x = 4\pi + (-1)^4(\pi/6) = 4\pi + \pi/6$. (Outside the interval)

4. Count the Solutions: There are 4 distinct solutions in the interval $[0, 3\pi]$.

Answer: There are 4 solutions.