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Trigonometric Equations Using Factoring

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Practice Trigonometric Equations Using Factoring
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Solving Trigonometric Equations Using Quadratic Techniques

As Agent Trigonometry, you are given this clue: 2cos2x3cosx+1=0\begin{align*}2\cos^2 x-3 \cos x + 1=0\end{align*} . If x\begin{align*}x\end{align*} falls in the interval 0x<2π\begin{align*}0 \le x < 2\pi\end{align*} , what is/are its possible value(s)?

Guidance

Another way to solve a trig equation is to use factoring or the quadratic formula. Let’s look at a couple of examples.

Example A

Solve sin2x3sinx+2=0\begin{align*}\sin^2x-3 \sin x+2=0\end{align*} .

Solution: This sine equation looks a lot like the quadratic x23x+2=0\begin{align*}x^2-3x+2=0\end{align*} which factors to be (x2)(x1)=0\begin{align*}(x-2)(x-1)=0\end{align*} and the solutions are x=2\begin{align*}x = 2\end{align*} and 1. We can factor the trig equation in the exact same manner. Instead of just x\begin{align*}x\end{align*} , we will have sinx\begin{align*}\sin x\end{align*} in the factors.

sin2x3sinx+2(sinx2)(sinx1)sinx=2 and sinx=0=0=1

There is no solution for sinx=2\begin{align*}\sin x=2\end{align*} and sinx=1\begin{align*}\sin x=1\end{align*} when x=π2±2πn\begin{align*}x= \frac{\pi}{2} \pm 2 \pi n\end{align*} .

Example B

Solve 1sinx=3cosx\begin{align*}1- \sin x=\sqrt{3} \cos x\end{align*} in the interval 0x<2π\begin{align*}0 \le x < 2 \pi\end{align*} .

Solution: To solve this equation, use the Pythagorean Identity sin2x+cos2x=1\begin{align*}\sin^2x+\cos^2x=1\end{align*} . Solve for either cosine and substitute into the equation. cosx=1sin2x\begin{align*}\cos x=\sqrt{1- \sin^2 x}\end{align*}

1sinx(1sinx)212sinx+sin2x4sin2x2sinx22sin2xsinx1(2sinx+1)(sinx1)=31sin2x=33sin2x2=33sin2x=0=0=0

Solving each factor for x\begin{align*}x\end{align*} , we get sinx=12x=7π6\begin{align*}\sin x=- \frac{1}{2} \rightarrow x=\frac{7 \pi}{6}\end{align*} and 11π6\begin{align*}\frac{11 \pi}{6}\end{align*} and sinx=1x=π2\begin{align*}\sin x=1 \rightarrow x=\frac{\pi}{2}\end{align*} .

Example C

Solve tan2x5tanx9=0\begin{align*}\tan^2x-5 \tan x-9=0\end{align*} in the interval 0x<π\begin{align*}0 \le x < \pi\end{align*} .

Solution: This equation is not factorable so you have to use the Quadratic Formula.

tanx=5±(5)24(1)(9)2=5±6126.41 and 1.41

xtan16.411.416 rad\begin{align*}x \approx \tan^{-1} 6.41 \approx 1.416 \ \text{rad}\end{align*} and xtan11.410.954 rad\begin{align*}x \approx \tan^{-1}-1.41 \approx -0.954 \ \text{rad}\end{align*}

The first answer is within the range, but the second is not. To adjust -0.954 to be within the range, we need to find the answer in the second quadrant, π0.954=2.186 rad\begin{align*}\pi - 0.954=2.186 \ \text{rad}\end{align*} .

Concept Problem Revisit We can solve this problem by factoring.

2cos2x3cosx+1=0(2cosx1)(cosx1)=0cosx=12ORx=1
.

Over the interval 0x<2π\begin{align*}0 \le x < 2\pi\end{align*} , cosx=12\begin{align*}\cos x = \frac{1}{2}\end{align*} when x=π3\begin{align*}x = \frac{\pi}{3}\end{align*} and cosx=1\begin{align*}\cos x = 1\end{align*} when x=0\begin{align*}x = 0\end{align*} .

Guided Practice

Solve the following trig equations using any method in the interval 0x<2π\begin{align*}0 \le x < 2 \pi\end{align*} .

1. sin2xcosx=cosx\begin{align*}\sin^2x \cos x=\cos x\end{align*}

2. sin2x=2sin(x)+1\begin{align*}\sin^2x=2 \sin(-x)+1\end{align*}

3. 4cos2x2cosx1=0\begin{align*}4 \cos^2x-2 \cos x-1=0\end{align*}

1. Put everything onto one side of the equation and factor out a cosine.

sin2xcosxcosxcosx(sin2x1)cosx(sinx1)(sinx+1)=0=0=0

cosxx=0  sinx=1  sinx=1=π2 and 3π2 x=π2x=3π2

2. Recall that sin(x)=sinx\begin{align*}\sin(-x)=- \sin x\end{align*} from the Negative Angle Identities.

sin2xsin2xsin2x+2sinx+1(sinx+1)2sinxx=2sin(x)+1=2sinx+1=0=0=1=3π2

cosx=2±224(4)(1)2(4)=2±208=1±254

xcos1(1+54)cos10.8090and0.6283xcos1(154)  cos10.3090  1.8850 (reference angle is π1.88501.2570)

The other solutions in the range are x2π0.62835.6549\begin{align*}x \approx 2 \pi - 0.6283 \approx 5.6549\end{align*} and xπ+1.25704.3982\begin{align*}x \approx \pi + 1.2570 \approx 4.3982\end{align*} .

Explore More

Solve the following trig equations using any method. Find all solutions in the interval 0x<2π\begin{align*}0 \le x < 2 \pi\end{align*} . Round any decimal answers to 4 decimal places.

1. 2cos2xsinx1=0\begin{align*}2 \cos^2x-\sin x -1=0\end{align*}
2. 4sin2x+5sinx+1=0\begin{align*}4 \sin^2x+5 \sin x+1=0\end{align*}
3. 3tan2xtanx=0\begin{align*}3 \tan^2x- \tan x=0\end{align*}
4. 2cos2x+cos(x)1=0\begin{align*}2 \cos^2x+\cos(-x)-1=0\end{align*}
5. 1sinx=2cosx\begin{align*}1- \sin x=\sqrt{2} \cos x\end{align*}
6. sinx=2sinx1\begin{align*}\sqrt{\sin x}=2 \sin x-1\end{align*}
7. sin3xsinx=0\begin{align*}\sin^3x-\sin x=0\end{align*}
8. tan2x8tanx+7=0\begin{align*}\tan^2x-8 \tan x+7=0\end{align*}
9. 5cos2x+3cosx2=0\begin{align*}5 \cos^2x+3 \cos x-2=0\end{align*}
10. sinxsinxcos2x=1\begin{align*}\sin x- \sin x \cos^2x=1\end{align*}
11. cos2x3cosx+2=0\begin{align*}\cos^2x-3 \cos x+2=0\end{align*}
12. sin2xcosx=4cosx\begin{align*}\sin^2x \cos x=4 \cos x\end{align*}
13. cosxcsc2x+2cosx=6cosx\begin{align*}\cos x \csc^2x+2 \cos x=6 \cos x\end{align*}

Using your graphing calculator, graph the following equations and determine the points of intersection in the interval 0x<2π\begin{align*}0 \le x < 2 \pi\end{align*} .

1. yy=sin2x=3sinx1
1. yy=4cosx3=2tanx