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Trigonometry : Solving Trigonometric Equations

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Example Questions

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Example Question #61 : Solving Trigonometric Equations

Find the roots for $y = \sin^2 x + 2 \sin x - 5$

Possible Answers:

90^o, 270^o

90^o

26.71^o, 153.29^o

60.56 ^o , 119.45^o

No solution

Correct answer:

No solution

Explanation:

To solve, use the quadratic formula:

$\sin \theta = \frac{ -2 \pm \sqrt{4 + 20 }}{2 } = \frac{-2 \pm \sqrt{24}}{2} = - \pm \sqrt{6}$

Both $-1 - \sqrt{6}$ and $-1+\sqrt{6}$ are outside of the range of the sine function, $-1 \leq \sin x \leq 1$ so there is no solution.

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Example Question #62 : Solving Trigonometric Equations

Solve for :

$0 = 2 \cos ^2 x + \cos x - 2$

Possible Answers:

38.67^o, 16.31^o

38.67^o, 321.33^o

141.33^o, 321.33^o

16.31^o, 343.69^o

38.67^o, 141.33^o

Correct answer:

38.67^o, 321.33^o

Explanation:

Solve using the quadratic formula:

$\cos x = \frac{-1 \pm \sqrt{1 + 16}}{4 } = \frac{-1 \pm \sqrt{17}}{4}$

$\frac{ -1 - \sqrt{17}}{4} \approx -1.28$ , outside the range for cosine.

$\frac{-1 + \sqrt{17}}{4} \approx 0.78$

$x = \cos ^{-1} (0.78 ) = 38.67^o$ according to a calculator.

The other angle with a cosine of 0.78 would be 360 ^o - 38.67^o = 321.33^o .

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Example Question #63 : Solving Trigonometric Equations

Solve for :

$0 = 4 \cos ^2 x - 21 \cos x + 5$

Possible Answers:

14.47^o, 165.52^o

14.47^o, 345.52^o

x = 75.52^o, 104.48^o

x = 75.52^o, 255.52 ^o

x = 75.52^o, 284.48^o

Correct answer:

x = 75.52^o, 284.48^o

Explanation:

Solve using the quadratic formula:

$\cos x = \frac{21 \pm \sqrt{441 - 4(4)(5)}}{2(4) } = \frac{ 21 \pm \sqrt{361}}{8} = \frac{21 \pm 19}{8}$

$\cos x = 5, \frac{1}{4}$

5 is outside the range for cosine, so the only solution that works is $\cos x = \frac{1}{4}$ :

$x = \cos ^{-1} (\frac{1}{4}) \approx 75.52 ^o$ according to a calculator

The other angle with a cosine of $\frac{1}{4}$ is 360^o - 75.52^o = 284.48^o

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Example Question #64 : Solving Trigonometric Equations

Solve for : $0 = 3 \cos ^2 x + 8 \cos x + 4$

Possible Answers:

131.81^o, 311.81^o

48.19^o, 228.19^o

48.19^o, 311.81^o

48.19^o, 131.81^o

131.81^o, 228.19^o

Correct answer:

131.81^o, 228.19^o

Explanation:

Use the quadratic formula:

$\cos x = \frac{ -8 \pm \sqrt{64 - 4(3)(4)}}{2(3)} = \frac{-8 \pm \sqrt{16}}{6 } = \frac{-8\pm4}{6}$

$\cos x = -2 \enspace or - \frac{2}{3}$

-2 is outside the range of cosine, so the answer has to come from $- \frac{2}{3}$ :

$\cos x = - \frac{2}{3}$

$x = \cos^{-1} (-\frac{2}{3} ) \approx 131.81^o$ according to a calculator

The other angle with a cosine of $-\frac{2}{3}$ is 360^o - 131.81^o = 228.19^o

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Example Question #61 : Solving Trigonometric Equations

Solve the equation

2cos^2x+sinx=1

for $0\leq x\leq2\pi$ .

Possible Answers:

$\frac{\pi}{2},\frac{\pi}{6}, \frac{2\pi}{3}$

$\frac{\pi}{2},\frac{\pi}{6}$

$\frac{\pi}{2}$

$\frac{7\pi}{6},\frac{11\pi}{6}$

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

Correct answer:

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

Explanation:

First of all, we can use the Pythagorean identity sin^2x+cos^2x=1 to rewrite the given equation in terms of sinx .

2cos^2x+sinx=1

$\Rightarrow 2(1-sin^2x)+sinx=1$

$\Rightarrow 2-2sin^2x+sinx=1$

$\Rightarrow -2sin^2x+sinx+1=0$

$\Rightarrow 2sin^2x-sinx-1=0$

This is a quadratic equation in terms of sinx ; hence, we can use the quadratic formula to solve this equation for sinx .

$sinx=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where a=2, b=-1, c=-1 .

$\Rightarrow sinx=\frac{1\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

$\Rightarrow sinx=\frac{1\pm3}{4}$

$\Rightarrow sinx=-\frac{1}{2}, sinx=1$ .

Now, sinx=1 when $x=\frac{\pi}{2}$ , and $sinx=-\frac{1}{2}$ when $x=\frac{7\pi}{6}$ or $\frac{11\pi}{6}$ .

Hence, the solutions to the original equation 2cos^2x+sinx=1 are

$\frac{\pi}{2},\frac{7\pi}{6}, \frac{11\pi}{6}$

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Example Question #61 : Solving Trigonometric Equations

In the interval $0\leq x\leq\pi$ , what values of x satisfy the following equation?

$1-\sin x=\cos^2x-\frac{1}{4}$

Possible Answers:

$x=\frac{\pi}{6},\frac{5\pi}{6}$

$x=\frac{\pi}{4},\frac{7\pi}{12}$

$x=\frac{\pi}{3},\frac{2\pi}{3}$

$x=0,\pi$

Correct answer:

$x=\frac{\pi}{6},\frac{5\pi}{6}$

Explanation:

We start by rewriting the $\cos^2x$ term on the right hand side in terms of $\sin^2x$ .

$\cos^2x=1-\sin^2x$

$1-\sin x=1-\sin^2x-\frac{1}{4}$

We then move everything to the left hand side of the equation and cancel.

$1-\sin x-1+\sin^2x+\frac{1}{4}=0$

$\sin^2x-\sin x+\frac{1}{4}=0$

Apply the quadratic formula:

$\begin{align*} \sin x&=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(\frac{1}{4})}}{2(1)}\\ &=\frac{1\pm\sqrt{1-1}}{2}\\ &=\frac{1}{2} \end{align*}$

So $\sin x=\frac{1}{2}$ . Using the unit circle, the two values of that yield this are $x=\frac{\pi}{6}$ and $x=\frac{5\pi}{6}$ .

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Trigonometry : Solving Trigonometric Equations

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #61 : Solving Trigonometric Equations

Example Question #62 : Solving Trigonometric Equations

Example Question #63 : Solving Trigonometric Equations

Example Question #64 : Solving Trigonometric Equations

Example Question #61 : Solving Trigonometric Equations

Example Question #61 : Solving Trigonometric Equations

All Trigonometry Resources

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