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Trigonometry : Trigonometric Equations

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6 Diagnostic Tests 155 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #81 : Trigonometric Equations

Solve for $\theta$ :

$y = 2 \cos ^2 (3 \theta ) + 5 \cos ( 3 \theta ) - 3$

Possible Answers:

$\frac{ 2 \pi }{ 9 } , \frac{ 4 \pi }{ 9 }$

no solution

$2 \pi$

$\frac{ \pi }{ 9 } , \frac{ 5 \pi }{9 }$

$\pi$

Correct answer:

$\frac{ \pi }{ 9 } , \frac{ 5 \pi }{9 }$

Explanation:

Use the quadratic formula to solve for $\cos (3 \theta )$ :

$\cos (3 \theta ) = \frac{ -5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{ -5 \pm \sqrt{49}}{4} = \frac{ -5 \pm 7 }{4}$

One possible solution is:

$\cos (3 \theta ) = \frac{- 5 - 7 }{4} = \frac{-12}{4} = -3$ this is outside of the possible range for cosine

The other solution is:

$\cos (3 \theta ) = \frac{ -5 + 7 }{4} = \frac{ 2}{4} =\frac{ 1}{2}$

$3 \theta = \cos ^ {-1 } (\frac{1}{2} ) = \frac{ \pi }{3} , \frac{ 5 \pi }{3}$ divide by 3

$\theta = \frac{ \pi }{ 9 } , \frac{ 5 \pi }{ 9 }$

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Example Question #241 : Trigonometry

Solve for $\theta$ :

$0 = 4 \sin ^ 4 \theta + 9 \sin ^ 2 \theta - 9$

Possible Answers:

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

$\frac{ \pi }{3} , \frac{ 5 \pi }{3}$

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

$\frac{ \pi }{6} , \frac{ 5 \pi }{6} , \frac{ 7 \pi }{6} , \frac{ 11 \pi }{6}$

$\frac{ \pi }{3} , \frac{ 2 \pi }{3} , \frac{ 4 \pi }{3} , \frac{ 5 \pi }{3}$

Correct answer:

$\frac{ \pi }{3} , \frac{ 2 \pi }{3} , \frac{ 4 \pi }{3} , \frac{ 5 \pi }{3}$

Explanation:

Solve using the quadratic formula:

$\sin ^ 2 \theta = \frac{ -9 \pm \sqrt{ 81 - 4(4)(-9)}}{2(4)} = \frac{ -9 \pm \sqrt{225}}{8 } = \frac{ -9 \pm 15}{8}$

One possible answer is:

$\sin ^ 2 \theta = \frac{ - 9 + 15 }{ 8 } = \frac{ 6 }{ 8 } = \frac{ 3}{4}$ take the square root

$\sin \theta = \pm \sqrt{ \frac{ 3}{4} } = \pm \frac{ \sqrt3}{2}$

$\theta = \sin ^ {-1} ( \pm \frac{ \sqrt3}{2} )$

$\theta = \frac{ \pi }{3} , \frac{2 \pi }{3} , \frac{ 4 \pi }{3} , \frac{ 5 \pi }{3}$

The other would be:

$\sin ^ 2 \theta = \frac{ -9 - 15 }{8 } = \frac{ - 24}{8 } = -3$ this is outside of the range for sine

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Example Question #51 : Solving Trigonometric Equations

Solve for $\theta$ : $5 = 2 \cos ^2 \theta - 5 \cos \theta + 2$

Possible Answers:

$\theta = \frac{2 \pi }{3} , \frac{4 \pi }{3}$

$\theta = \frac{ \pi }{3}$

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

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

$\theta = \frac{ 2 \pi }{3}$

Correct answer:

$\theta = \frac{2 \pi }{3} , \frac{4 \pi }{3}$

Explanation:

Subtracting 5 from both sides gives the quadratic equation $0 = 2 \cos ^2 \theta - 5 \cos \theta - 3$

Using the quadratic formula gives:

$\cos \theta = \frac{ 5 \pm \sqrt{25 - 4(2)(-3)}}{2(2)} = \frac{ 5 \pm 7}{4} = \frac{ -1}{2} , 3$

The cosine cannot be 3 because that's greater than 1.

$\cos \theta = - \frac{1}{2}$

$\theta = \cos ^{-1} (-\frac{1}{2}) = \frac{2 \pi }{3} , \frac{ 4 \pi }{3}$

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Example Question #241 : Trigonometry

Which is not a solution for $\theta$ for $10 \sin ^2 \theta + \sin \theta - 3 = 0$ ?

Possible Answers:

150^o

143.13^o

60^o

323.13^o

30^o

Correct answer:

60^o

Explanation:

Using the quadratic formula gives:

$\sin \theta = \frac{ -1 \pm \sqrt{ 1 - 4(10)(-3)}}{2(10)} = \frac{-1 \pm 11}{20 } = \frac{1}{2} , -\frac{3}{5}$

$\theta = \sin ^{-1} (\frac{ 1}{2} ) = 30^o, 150^o$ or

$\theta = \sin ^{-1} (-\frac{3}{5}) \approx 323.13^o, 143.13^o$

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Example Question #83 : Trigonometric Equations

Solve for $\theta$ : $0=2 \sin ^2 \theta - 3 \sin \theta + 1$

Possible Answers:

$\frac{\pi }{2} , \frac{3 \pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

$\frac{\pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

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

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

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

Correct answer:

$\frac{\pi }{2} , \frac{\pi }{6} , \frac{5 \pi }{6}$

Explanation:

Solve using the quadratic formula:

$\sin \theta = \frac{ 3 \pm \sqrt{9 - 4(2)}}{4(1)} = \frac{ 3 \pm \sqrt1}{4} = \frac{1}{2} \enspace or \enspace 1$

$\sin \theta = \frac{1}{2}$

$\theta = \sin ^{-1} (\frac{1}{2} ) = \frac{\pi }{6} , \frac{ 5 \pi }{6}$

$\sin \theta = 1$

$\theta = \sin ^{-1}(1)$

$\theta = \frac{\pi }{2}$

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Example Question #21 : Quadratic Formula With Trigonometry

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

Possible Answers:

60.56 ^o , 119.45^o

26.71^o, 153.29^o

No solution

90^o, 270^o

90^o

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 #22 : Quadratic Formula With Trigonometry

Solve for :

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

Possible Answers:

38.67^o, 141.33^o

141.33^o, 321.33^o

16.31^o, 343.69^o

38.67^o, 321.33^o

38.67^o, 16.31^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 #23 : Quadratic Formula With Trigonometry

Solve for :

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

Possible Answers:

x = 75.52^o, 255.52 ^o

14.47^o, 345.52^o

x = 75.52^o, 104.48^o

x = 75.52^o, 284.48^o

14.47^o, 165.52^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 #24 : Quadratic Formula With Trigonometry

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

Possible Answers:

48.19^o, 131.81^o

131.81^o, 228.19^o

48.19^o, 311.81^o

131.81^o, 311.81^o

48.19^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 #251 : Trigonometry

Solve the equation

2cos^2x+sinx=1

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

Possible Answers:

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

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

$\frac{\pi}{2}$

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

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

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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Trigonometry : Trigonometric Equations

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #81 : Trigonometric Equations

Example Question #241 : Trigonometry

Example Question #51 : Solving Trigonometric Equations

Example Question #241 : Trigonometry

Example Question #83 : Trigonometric Equations

Example Question #21 : Quadratic Formula With Trigonometry

Example Question #22 : Quadratic Formula With Trigonometry

Example Question #23 : Quadratic Formula With Trigonometry

Example Question #24 : Quadratic Formula With Trigonometry

Example Question #251 : Trigonometry

All Trigonometry Resources

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