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

Solving Quadratic Trigonometric Equations

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Solving Quadratic Trigonometric Equations

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

How to Solve Quadratic Trigonometric Equations

Let af2(θ)+bf(θ)+c=0af^2(\theta)+bf(\theta)+c=0 be a quadratic trigonometric equation where f(θ)f(\theta) is a trigonometric function and a,b,c∈Ra,b,c\in\mathbb{R}.
Solve for θ\theta using the following steps:
  1. Factor:
af2(θ)+bf(θ)+c=(f(θ)−A)(f(θ)−B)af^2(\theta)+bf(\theta)+c=(f(\theta)-A)(f(\theta)-B) where A, B∈RA,~B\in\mathbb{R}.
  1. Solve for θ\theta:
f(θ)=A, Bθ=f−1(A), f−1(B)\begin{array}{rcl} f(\theta)&=&A,~B\\\\ \theta&=&f^{-1}(A),~f^{-1}(B) \end{array}


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Example

Let tan⁡2θ+tan⁡θ−2=0\tan^2\theta+\tan\theta-2=0. Solve for θ\theta over the domain 0≤θ≤2π0\leq{}\theta\leq{}2\pi. Round to the nearest thousandth.


tan⁡2θ+tan⁡θ−2=0(tan⁡θ+2)(tan⁡θ−1)=0∴ tan⁡θ=−2&tan⁡θ=1\begin{array}{rcl} \tan^2\theta+\tan\theta-2&=&0\\\\ (\tan\theta+2)(\tan\theta-1)&=&0\\\\ \therefore~\tan\theta=-2&\&&\tan\theta=1 \end{array}

tan⁡θ=−2:‾\colorOne{\underline{\tan\theta=-2:}}

Since −2-2 is not a special side length, then we must use a graphing calculator,

Find the reference angle

(For a solution without the graphing calculator, see the video.)

In a graphing calculator, press 2nd  +  TAN  +  2\boxed{2^{\text{nd}}}~~+~~\boxed{\text{TAN}}~~+~~\boxed{2} .

Therefore, θRA=1.107\colorFive{\theta_{RA}=1.107}

Since tan⁡θ=−2<0\tan\theta=-2<0, then θ\theta is in quadrant II & IV.

θ1=π−1.107≈2.034θ2=2π−1.107≈5.176\begin{array}{rcl} \theta_{1}&=&\pi-1.107&\approx&\boxed{2.034}\\\\ \theta_{2}&=&2\pi-1.107&\approx&\boxed{5.176} \end{array}


tan⁡θ=1:‾\colorOne{\underline{\tan\theta=1:}}

Since tan⁡θ=sin⁡θcos⁡θ=1\tan\theta=\dfrac{\sin\theta}{\cos\theta}=1, then sin⁡θ=cos⁡θ\sin\theta=\cos\theta.

∴ θ=π4, 5π4\therefore~\theta=\boxed{\dfrac{\pi}{4},~\dfrac{5\pi}{4}}

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Example: Solving Quadratic Trigonometric Equations

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

2sin⁡θ=3+2csc⁡θ2\sin\theta=3+2\csc\theta



2sin⁡θ=3+2csc⁡θ2sin⁡θ=3+2sin⁡θ2sin⁡2θ=3sin⁡θ+22sin⁡2θ−3sin⁡θ−2=0(2sin⁡2θ−4sin⁡θ)+(sin⁡θ−2)=02sin⁡θ(sin⁡θ−2)+1(sin⁡θ−2)=0(sin⁡θ−2)(2sin⁡θ+1)=0∴ sin⁡θ=2&sin⁡θ=−12\begin{array}{rcl} 2\sin\theta&=&3+2\csc\theta\\\\ 2\sin\theta&=&3+\dfrac{2}{\sin\theta}\\\\ 2\sin^2\theta&=&3\sin\theta+2\\\\ 2\sin^2\theta-3\sin\theta-2&=&0\\\\ (2\sin^2\theta-4\sin\theta)+(\sin\theta-2)&=&0\\\\ 2\sin\theta(\sin\theta-2)+1(\sin\theta-2)&=&0\\\\ (\sin\theta-2)(2\sin\theta+1)&=&0\\\\ \therefore~\sin\theta=2&\&&\sin\theta=-\dfrac{1}{2} \end{array}

sin⁡θ=2:‾\colorOne{\underline{\sin\theta=2:}}

There are no real values of θ\theta that satisfy this equation.


sin⁡θ=−12:‾\colorOne{\underline{\sin\theta=-\frac{1}{2}:}}

Find the reference angle:

sin⁡θ=−12θRA=sin⁡−1(∣−12∣)θRA=π6\begin{array}{rcl} \sin\theta&=&-\dfrac{1}{2}\\\\ \theta_{RA}&=&\sin^{-1}\Bigg(\Bigg|-\dfrac{1}{2}\Bigg|\Bigg)\\\\ \theta_{RA}&=&\dfrac{\pi}{6} \end{array}


Since sin⁡θ=−12<0\sin\theta=-\dfrac{1}{2}<0, then θ\theta is in quadrant III & IV.

θ1=π+π6=7π6θ2=2π−π6=11π6\begin{array}{rcl} \theta_{1}&=&\pi+\dfrac{\pi}{6}&=&\boxed{\dfrac{7\pi}{6}}\\\\ \theta_{2}&=&2\pi-\dfrac{\pi}{6}&=&\boxed{\dfrac{11\pi}{6}} \end{array}

Practice: Solving Quadratic Trigonometric Equations

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

3sec⁡2θ−4=03\sec^2\theta-4=0
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Practice: Solving Quadratic Trigonometric Equations

Let 2tan⁡2θ=3tan⁡θ−12\tan^2\theta=3\tan\theta-1. Find θ\theta over the domain 0≤θ≤2π0\le\theta\le2\pi.



Practice: Solving Quadratic Trigonometric Equations

Let 4sin⁡θcos⁡θ−1=04\sin\theta\cos\theta-1=0. Find θ\theta and the general solutions.




Extra Practice
Solving Quadratic Trigonometric Equations
Practice: Solving Quadratic Trigonometric Equations
Let 2tan⁡2θ=3tan⁡θ−12\tan^2\theta=3\tan\theta-1. Find θ\theta over the domain 0≤θ≤2π0\le\theta\le2\pi.
Solving Quadratic Trigonometric Equations
Practice: Solving Quadratic Trigonometric Equations
Let 4sin⁡θcos⁡θ−1=04\sin\theta\cos\theta-1=0. Find θ\theta and the general solutions.
Solving Quadratic Trigonometric Equations
Practice: Solving Quadratic Trigonometric Equations
Solve for θ\theta over the domain 0≤θ≤2π0\leq{}\theta\leq{}2\pi .
3sec⁡2θ−4=03\sec^2\theta-4=0
Trigonometric Identities & Proofs
Solve for xx over the domain 0≤x<2π0\leq{}x<2\pi.
cos⁡2x=sin⁡x\cos{2x}=\sin{x}
Solving Quadratic Trigonometric Equations
Solve for θ\theta over the domain −2π≤θ<2π-2\pi\leq\theta<2\pi.
2sin⁡2θ−sin⁡θ−32\sin^2\theta-\sin\theta-3