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Wize High School Grade 12 Pre-Calculus Textbook > Trigonometric Ratios (Radians)

Solving Simple Trigonometric Equations

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Solving Simple Trigonometric Equations

A simple trigonometric equation is in the form:
f(θ)=kf(\theta)=k

where k∈Rk\in\mathbb{R} and f(θ)f(\theta) is a trigonometric function.

Step 1.
Identify what quadrant(s) the angle lies in.

Step 2.
Identify the reference angle.

Step 3.
Solve for the solutions in the appropriate quadrants.

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Example 1

Solve for θ\theta over the domain 0≤θ<2π0\leq{\theta}<2\pi.

cos⁡θ=12\cos{\theta}=\displaystyle\frac{1}{2}

Step 1.
Identify what quadrant(s) the angle lies in.

Since cos⁡θ>0\cos{\theta}>0, then cos⁡θ\cos{\theta} is in quadrant I & IV.

Step 2.
Identify the reference angle.

Since cos⁡θ=x\cos{\theta}=x, then we need the angle that corresponds to an x-coordinate point of 12\displaystyle\frac{1}{2}.
The reference angle is π3\displaystyle\frac{\pi}{3}.

Step 3.
Solve for the solutions in the appropriate quadrants.

Quadrant I:

π3\displaystyle\frac{\pi}{3}

Quadrant IV:

2π−π3=5π32\pi-\displaystyle\frac{\pi}{3}=\displaystyle\frac{5\pi}{3}

The solutions are θ=π3, 5π3\theta=\displaystyle\frac{\pi}{3},~\displaystyle\frac{5\pi}{3}
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Example: Solving Simple Trigonometric Equations


Solve cos⁡θ=−32\cos{\theta}=\displaystyle-\frac{\sqrt{3}}{2} :

a. Over the domain 0≤θ<360∘0\leq{}\theta<360^\circ

b. Provide the general solution


a.

Step 1.

Since cos⁡θ<0,\cos{\theta}<0, then θ\theta is in quadrant II & III.

Step 2.

The reference angle is π6\displaystyle\frac{\pi}{6}

Step 3.

Quadrant II: π−π6=5π6\pi-\displaystyle\frac{\pi}{6}=\displaystyle\frac{5\pi}{6}

Quadrant III; π+π6=7π6\pi+\displaystyle\frac{\pi}{6}=\displaystyle\frac{7\pi}{6}

b.
The general solution is all coterminal angles to the solutions in Step 3.

Therefore,

5π6±2nπ,  n∈W\displaystyle\frac{5\pi}{6}\pm2n\pi,~~n\in\mathbb{W}

7π6±2nπ,  n∈W\displaystyle\frac{7\pi}{6}\pm2n\pi,~~n\in\mathbb{W}
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Example: Solving Simple Trigonometric Equations


Solve csc⁡θ=−2\csc{\theta}=-\sqrt{2} :

a. Over the domain 0≤θ<360°0\le\theta<360^{\degree }

b. Provide the general solution


a.

Step 1.

Since csc⁡θ<0,\csc{\theta}<0, then θ\theta is in quadrant III & IV.

Step 2.

The reference angle is π4\displaystyle\frac{\pi}{4}

Step 3.

Quadrant II: π+π4=5π4\pi+\displaystyle\frac{\pi}{4}=\displaystyle\frac{5\pi}{4}

Quadrant III; 2π−π4=11π42\pi-\displaystyle\frac{\pi}{4}=\displaystyle\frac{11\pi}{4}

b.
The general solution is all coterminal angles to the solutions in Step 3.

Therefore,

5π4±2nπ,  n∈W\displaystyle\frac{5\pi}{4}\pm2n\pi,~~n\in\mathbb{W}

11π4±2nπ,  n∈W\displaystyle\frac{11\pi}{4}\pm2n\pi,~~n\in\mathbb{W}

Practice: Solving Simple Trigonometric Equations

Solve cot⁡θ=0\cot{\theta}=0 over the domain 0≤θ<2π0\leq{}\theta<2\pi.




Practice: Solving Simple Trigonometric Equations


Let 3cos⁡θ+23cos⁡θ=27\sqrt{3}\cos{\theta}+2\sqrt{3}\cos{\theta}=\sqrt{27}. Solve for θ \theta~ over the domain 0≤θ<2π0\leq{\theta}<2\pi.




Practice: Solving Simple Trigonometric Equations

Let sin⁡2θ=12\sin{2\theta}=\displaystyle\frac{1}{2}. Solve for θ\theta, providing the general solution.



Extra Practice
Solving Trigonometric Equations
Find aa over the domain 0≤a<2π0\leq{}a<2\pi
cot⁡a2=1\cot\dfrac{a}{2}=1
Solving Simple Trigonometric Ratios
Solve for θ\theta over the interval (−∞,∞)(-\infin,\infin):
4cos⁡3θ=24\cos3\theta=2
Solving Simple Trigonometric Equations
Solve for θ\theta over the domain 0≤θ<2π0\leq{}\theta<2\pi
2sin⁡θ=−22\sin\theta=-\sqrt{2}
Solving Simple Trigonometric Equations
Practice: Solving Simple Trigonometric Equations
Let tan⁡2θ=1\tan{2\theta}=1. Solve for θ\theta, providing the general solution.
Solving Simple Trigonometric Equations
Practice: Solving Simple Trigonometric Equations
Let cos⁡2θ=−32\cos{2\theta}=-\dfrac{\sqrt{3}}{2}. Solve for θ\theta, providing the general solution.
Solving Simple Trigonometric Equations
Practice: Solving Simple Trigonometric Equations
Let 32sec⁡θ−2sec⁡θ=43\sqrt{2}\sec{\theta}-\sqrt{2}\sec{\theta}=4. Solve for θ \theta~ over the domain 0≤θ<2π0\leq{\theta}<2\pi.
Solving Simple Trigonometric Equations
Practice: Solving Simple Trigonometric Equations
Let 4sin⁡θ−2sin⁡θ=14\sin\theta-2\sin\theta=1. Solve for θ \theta~ over the domain 0≤θ<2π0\leq{\theta}<2\pi.
Solving Simple Trigonometric Equations
Practice: Solving Simple Trigonometric Equations
Solve cos⁡θ=12\cos{\theta}=\dfrac{1}{\sqrt{2}} over the domain 0≤θ<2π0\leq{}\theta<2\pi.
Solving Simple Trigonometric Equations
Practice: Solving Simple Trigonometric Equations
Solve sin⁡θ=−32\sin{\theta}=-\dfrac{\sqrt{3}}{2} over the domain 0≤θ<2π0\leq{}\theta<2\pi.