Principal and General Solutions

Introduction to Trigonometric Equations

An equation involving one or more trigonometric functions of an unknown angle is called a trigonometric equation. For example, $\sin x = 1/2$ is a trigonometric equation. The solutions to these equations are the values of the unknown angle 'x' that make the equation true.

Principal Solutions

Due to the periodic nature of trigonometric functions, there are infinite solutions to most trigonometric equations. The solutions that lie within the interval $[0, 2\pi)$ (one full revolution) are called principal solutions.

• Example: For the equation $\sin x = 1/2$, the angle in the first quadrant is $x=\pi/6$. Sine is also positive in the second quadrant, giving another solution $x = \pi - \pi/6 = 5\pi/6$. Both $\pi/6$ and $5\pi/6$ are in the interval $[0, 2\pi)$, so they are the principal solutions.

General Solutions

The expression involving an integer 'n' which gives all possible solutions of a trigonometric equation is called the general solution. This accounts for the periodicity of the functions by adding integer multiples of the period.

Formulas for General Solutions:

Let $\alpha$ be the principal value (usually the smallest positive angle) that satisfies the equation.

• If $\sin x = \sin \alpha$, the angles have the same y-coordinate on the unit circle. This occurs at $\alpha$ and $\pi-\alpha$, plus any number of full rotations. This is concisely written as:

$$x = n\pi + (-1)^n \alpha, \quad n \in \mathbb{Z}$$

• If $\cos x = \cos \alpha$, the angles have the same x-coordinate. This occurs at $\alpha$ and $-\alpha$, plus any number of full rotations.

$$x = 2n\pi \pm \alpha, \quad n \in \mathbb{Z}$$

• If $\tan x = \tan \alpha$, the angles are separated by half-revolutions, as the period of tangent is $\pi$.

$$x = n\pi + \alpha, \quad n \in \mathbb{Z}$$

General Solutions for Squared Functions:

• If $\sin^2x = \sin^2\alpha$, $\cos^2x = \cos^2\alpha$, or $\tan^2x = \tan^2\alpha$, the solutions occur in all four quadrants with the same reference angle $\alpha$. This pattern is captured by a single formula:

$$x = n\pi \pm \alpha, \quad n \in \mathbb{Z}$$

Example 1: Finding Principal Solutions

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

Solution:

Step 1: Identify the base angle. We know that $\sin(\pi/3) = \sqrt{3}/2$. This is our first solution.

Step 2: Use the ASTC rule to find other solutions. Sine is positive in the first and second quadrants.

The angle in the second quadrant with the same reference angle is $\pi - \pi/3 = 2\pi/3$.

Step 3: Check if the solutions are in the interval $[0, 2\pi)$. Both $\pi/3$ and $2\pi/3$ are in this interval. The principal solutions are $\pi/3$ and $2\pi/3$.

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Example 2: Finding General Solution for Cosine

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

Solution:

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

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

Step 3: Apply the general solution formula for cosine, $x = 2n\pi \pm \alpha$.

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

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Example 3: Finding General Solution for Tangent

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

Solution:

Step 1: Convert both sides to the same function. We use the identity $-\cot(\theta) = \tan(\pi/2+\theta)$.

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

Step 2: This is of the form $\tan A = \tan B$. The general solution is $A = n\pi + B$.

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

Step 3: Solve for x.

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

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Example 4: Finding General Solution for Sine

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

Solution:

Step 1: Find the principal value $\alpha$. Sine is negative in Q3 and Q4. The reference angle for $\sin \theta = \sqrt{3}/2$ is $\pi/3$. The smallest positive angle is in Q3, so $\alpha = \pi + \pi/3 = 4\pi/3$. The equation becomes $\sin(2x) = \sin(4\pi/3)$.

Step 2: Use the general solution formula for sine, $A = n\pi + (-1)^n B$.

$2x = n\pi + (-1)^n (4\pi/3)$.

Step 3: Solve for x.

$x = \mathbf{\frac{n\pi}{2} + (-1)^n \frac{2\pi}{3}, n \in \mathbb{Z}}$.

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Example 5: Solving a Quadratic in a Trig Function

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

Solution:

Step 1: Express the equation in terms of a single trigonometric function. Use the identity $\sin^2x = 1-\cos^2x$.

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

Step 2: Solve the quadratic. Let $y=\cos x$. The equation is $2y^2-3y-2=0$. Factoring gives $(2y+1)(y-2)=0$. So, $y=-1/2$ or $y=2$.

Step 3: Reject invalid solutions. The range of $\cos x$ is $[-1,1]$, so $\cos x=2$ is not possible.

Step 4: Find the general solution for the valid case. We solve $\cos x = -1/2$. From Example 2, the general solution is $x = 2n\pi \pm \frac{2\pi}{3}, n \in \mathbb{Z}$.

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Example 6: Solving with Squared Functions

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

Solution:

Step 1: The equation is in the form $\tan^2x = C$. We find the principal angle $\alpha$ such that $\tan^2\alpha=3$. Taking the square root, $\tan\alpha = \sqrt{3}$, which gives $\alpha = \pi/3$.

Step 2: The equation can be written as $\tan^2x = \tan^2(\pi/3)$.

Step 3: Use the general solution formula for squared functions, $x = n\pi \pm \alpha$.

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

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Example 7: Equation with sec and tan

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

Solution:

Step 1: Rewrite in terms of sine and cosine: $\frac{1}{\cos x} - \frac{\sin x}{\cos x} = \sqrt{3} \implies 1-\sin x = \sqrt{3}\cos x$ (provided $\cos x \neq 0$).

Step 2: Rearrange into the form $a\cos x + b\sin x = c$. This gives $\sqrt{3}\cos x + \sin x = 1$. Divide by $\sqrt{a^2+b^2}=\sqrt{(\sqrt{3})^2+1^2}=2$.

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

Step 3: Convert the LHS to a single trigonometric function. This matches the form $\sin(A+B) = \sin A \cos B + \cos A \sin B$. It is $\sin(x+\pi/3) = 1/2$.

Step 4: Find the general solution. $\sin(x+\pi/3) = \sin(\pi/6)$.

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

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