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Wize High School Grade 12 Pre-Calculus Textbook > Trigonometric Equations

Solving Linear Trigonometric Equations

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Solving Linear Trigonometric Equations


Solving linear trigonometric equations is a very similar process to solving linear equations but with a few extra steps!

How to Solve a Linear Trigonometric Equation

Let af(θ)−b=0af(\theta)-b=0 be a linear trigonometric equation where f(θ)f(\theta) is a trigonometric function and a,b∈Ra,b\in\mathbb{R}.
Solve for θ\theta using the following steps:
  1. Isolate f(θ)f(\theta) by letting f(θ)=baf(\theta)=\dfrac{b}{a}.
  2. Find the reference angle θRA=f−1(∣ba∣)\theta_{\text{RA}}=f^{-1}\Bigg(\Bigg|{\dfrac{b}{a}\Bigg|}\Bigg)
  3. Determine which quadrants θ\theta lies in.
  4. Using the reference angle, find the angles in the appropriate quadrants.

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Example
Let 2cos⁡θ=12\cos\theta=1. Solve for θ\theta over the domain 0≤θ≤2π0\leq{}\theta\leq{}2\pi.

Step 1/2.

2cos⁡θ=1cos⁡θ=12θRA=cos⁡−1(12)θRA=π3\begin{array}{rcl} 2\cos\theta&=&1\\\\ \cos\theta&=&\dfrac{1}{2}\\\\ \theta_{\text{RA}}&=&\cos^{-1}\Bigg(\dfrac{1}{2}\Bigg)\\\\ \theta_{\text{RA}}&=&\dfrac{\pi}{3} \end{array}

Step 3.

Since cos⁡θ=12>0\cos\theta=\dfrac{1}{2}>0, then θ\theta lies in quadrants I & IV.

Step 4.

Since the solution lies in quadrants I & IV, then our solutions are:

θ1=π3\theta_{1}=\boxed{\dfrac{\pi}{3}}

θ2=2π−π3=5π3\begin{array}{rcl} \theta_2&=&2\pi-\dfrac{\pi}{3}\\\\ &=&\boxed{\dfrac{5\pi}{3}} \end{array}
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Example: Solving Linear Trigonometric Equations


Let 3tan⁡θ+1=0\sqrt{3}\tan\theta+1=0. Solve for θ\theta, providing the general solutions.


3tan⁡θ+1=0tan⁡θ=−13θRA=tan⁡−1(∣−13∣)θRA=π6\begin{array}{rcl} \sqrt{3}\tan\theta+1&=&0\\\\ \tan\theta&=&-\dfrac{1}{\sqrt{3}}\\\\ \theta_{RA}&=&\tan^{-1}\Bigg(\Bigg|-\dfrac{1}{\sqrt{3}}\Bigg|\Bigg)\\\\ \theta_{RA}&=&\dfrac{\pi}{6} \end{array}


Since tan⁡θ=−13<0,\tan\theta=-\dfrac{1}{\sqrt{3}}<0, then θ\theta is in quadrant II & IV.

Therefore, the solutions are:

θ1=π−π6=5π6+2nπ, n∈Rθ2=2π−π6=11π6+2nπ, n∈R\begin{array}{rclcl} \theta_{1}&=&\pi-\dfrac{\pi}{6}&=&\boxed{\dfrac{5\pi}{6}+2n\pi,~n\in\mathbb{R}}\\\\ \theta_{2}&=&2\pi-\dfrac{\pi}{6}&=&\boxed{\dfrac{11\pi}{6}+2n\pi,~n\in\mathbb{R}} \end{array}

Practice: Solving Linear Trigonometric Functions


Solve csc⁡θ=±2\csc\theta=\pm\sqrt{2} , providing the general solution.

Practice: Solving Linear 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 Linear Trigonometric Functions


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

cos⁡θtan⁡θ−3cos⁡θ−sin⁡θtan⁡θ+3sin⁡θ=0\cos\theta\tan\theta-\sqrt{3}\cos\theta-\sin\theta\tan\theta+\sqrt{3}\sin\theta=0




Extra Practice
Solving Linear Trigonometric Functions
Practice: Solving Linear Trigonometric Functions
Solve for θ\theta over the domain 0≤θ≤2π0\leq{}\theta\leq{}2\pi.
cos⁡θtan⁡θ−3cos⁡θ−sin⁡θtan⁡θ+3sin⁡θ=0\cos\theta\tan\theta-\sqrt{3}\cos\theta-\sin\theta\tan\theta+\sqrt{3}\sin\theta=0
Solving Linear Trigonometric Equations
Practice: Solving Linear 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.
Solving Linear Trigonometric Functions
Practice: Solving Linear Trigonometric Functions
Solve csc⁡θ=±2\csc\theta=\pm\sqrt{2} , providing the general solution.