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Trigonometry : Trigonometry

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

Example Questions

Example Question #51 : Trigonometric Equations

Find the roots of the equation $y = \frac{ \sqrt 3 }{3} - \frac{ 2}{3} \sin (\frac{ \theta }{2})$

Possible Answers:

$\frac{ \pi }{3} , \frac{ 11 \pi }{3}$

no solution

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

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

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

Correct answer:

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

Explanation:

There are many ways to solve this problem, and here is one.

First, re-write the equation so that it is set equal to zero, since we're finding the roots, and so that it is all in one fraction. That is easy to do since both terms already have the common denominator of 3:

$\frac{ \sqrt3 - 2 \sin (\frac{\theta }{2} ) } {3 } = 0$ multiply both sides by 3

$\sqrt3 - 2 \sin (\frac{ \theta }{ 2 }) = 0$ add the sine term to both sides

$\sqrt3 = 2 \sin (\frac{ \theta }{2 })$ divide both sides by 2

$\frac{ \sqrt3 }{2} = \sin (\frac{ \theta }{2} )$

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

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

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

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Example Question #52 : Trigonometric Equations

Find the roots of the equation $2 \cos ^2 (\frac{ 2 }{3} \theta ) \sin \theta - \sin \theta + 10 \cos ^2 (\frac{2}{3} \theta ) - 5 = y$

Possible Answers:

$\frac{ \pi }{4} , \frac{ 7 \pi }{4} , \frac{ 3 \pi }{4} , \frac{ 5 \pi }{4}$

$\frac{ 3 \pi }{8 } , \frac{ 21 \pi }{8 }$

$\frac{ \pi }{6} , \frac{ 7 \pi }{6} , \frac{ \pi }{2} , \frac{ 5 \pi }{6}$

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

$\frac{ 3 \pi }{8 } , \frac{ 5 \pi }{8 } , \frac{ 9 \pi }{8 } , \frac{ 15 \pi }{8 }$

Correct answer:

$\frac{ 3 \pi }{8 } , \frac{ 5 \pi }{8 } , \frac{ 9 \pi }{8 } , \frac{ 15 \pi }{8 }$

Explanation:

Factor the equation by grouping: the first 2 terms have $\sin \theta$ in common, and the second 2 have 5 in common:

$y = \sin \theta [ 2 \cos ^2 (\frac{ 2}{3 } \theta ) - 1 ] + 5 [ 2 \cos ^2 (\frac{2}{3} \theta ) - 1 ]$

$y = (2 \cos ^2 (\frac{ 2}{3} \theta ) - 1 ) ( \sin \theta + 5 )$

Since we are finding the roots, we can set each factor equal to 0:

$\sin \theta + 5 = 0$ subtract 5 from both sides

$\sin \theta = -5$ this is outside of the range for sine

$2 \cos ^2 (\frac{ 2}{3} \theta ) - 1 = 0$ add 1 to both sides

$2 \cos ^ 2 ( \frac{ 2}{3 } \theta ) = 1$ divide both sides by 2

$\cos ^2 ( \frac{ 2}{3 }\theta ) = \frac{ 1}{2}$ take the square root of both sides

$\cos (\frac{ 2}{3} \theta ) = \pm \frac{ 1 }{\sqrt2}$

$\frac{2}{3} \theta = \cos ^ {-1} ( \pm \frac{ 1}{ \sqrt2})$

$\frac{ 2}{3} \theta = \frac{ \pi }{4} , \frac{ 7 \pi }{4} , \frac{ 3 \pi }{4} , \frac{ 5 \pi }{4}$ multiply by $\frac{ 3}{2 }$

$\theta = \frac{ 3 \pi }{8 } , \frac{ 21 \pi }{8 } , \frac{ 9 \pi }{8 } , \frac{ 15 \pi }{8 }$

That second one is greater than $2\pi$ , so we want to figure out which angle between 0 and $2 \pi$ is coterminal with it. We can figure this out by subtracting $2 \pi$ , or in this case $\frac{ 16 \pi }{ 8 }$ to get $\frac{ 21 \pi }{ 8 } - \frac{ 16 \pi }{ 8 } = \frac{ 5 \pi } {8 }$

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Example Question #22 : Finding Trigonometric Roots

Find the roots of the equation $y= 2 \sin \theta \cos (4 \theta ) - \sqrt3 \cos (4 \theta ) + 2 \sin \theta - \sqrt3$

Possible Answers:

$\pi , 2 \pi$

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

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

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

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

Correct answer:

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

Explanation:

Factor by grouping. The first 2 terms have $\cos(4 \theta)$ in common.

$y = \cos (4\theta ) [ 2 \sin \theta - \sqrt3 ] + 1[2 \sin \theta - \sqrt3]$

$y = ( \cos (4 \theta ) + 1 ) ( 2 \sin \theta - \sqrt3 )$

We are finding the roots, so set each factor equal to 0:

$\cos (4 \theta ) + 1 = 0$ subtract 1 from both sides

$\cos (4 \theta ) = -1$

$4 \theta = \cos ^ {-1 } (-1)$

$4 \theta = \pi$ divide by 4

$\theta = \frac{ \pi }{4}$

$2 \sin \theta - \sqrt3 = 0$ add $\sqrt3$ to both sides

$2 \sin \theta = \sqrt 3$ divide both sides by 2

$\sin \theta = \frac{ \sqrt3}{2 }$

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

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

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Example Question #221 : Trigonometry

Find the roots of the equation $y = 2 \sin x \cos x - \sin x - 2 \cos x + 1$

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Factor the equation:

$y = \sin x (2 \cos x - 1 ) - 1 (2 \cos x - 1 )$ $y = (\sin x - 1 )(2 \cos x - 1 )$

The roots occur when each factor equals 0:

$\sin x - 1 = 0$

$\sin x = 1$

$x = \sin ^{-1}(1) = \frac{\pi }{2}$

$2 \cos x - 1 = 0$

$2 \cos x = 1$

$\cos x = \frac{1}{2}$

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

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Example Question #222 : Trigonometry

Find all roots of this equation, in radians.

$\tan^2 x=3/4$

Possible Answers:

x=0.64

x=-2.28

x=0.64, x=-0.64

x=0.71, x=-0.71

x=1.04, x=-1.04

Correct answer:

x=0.71, x=-0.71

Explanation:

Take the square root of both sides:

$\tan^2 x=3/4\rightarrow\tan x=\pm\sqrt{3}/2$

This gives us two roots,

$\tan x=\sqrt{3}/2$ and $\tan x=-\sqrt{3}/2$

Solving for x:

$x=\arctan (\sqrt{3}/2)$ or $x=\arctan (-\sqrt{3}/2)$ ,

which evaluate to x=0.71 and x=-0.71

Now we check to make sure both answers work. Make sure your calculator is in radians mode!

$\tan^2 (0.71)=0.74\approx 3/4$

$\tan^2 (-0.71)=0.74\approx 3/4$

Our answers are x=0.71, x=-0.71

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Example Question #223 : Trigonometry

If $\small \tan x * \cos x = 1$ What is in degrees?

Possible Answers:

Correct answer:

Explanation:

if $\small \tan x$ is $\small \frac{\sin x }{\cos x }$

$\frac{\sin(x)}{\cos(x)}\cdot \cos(x)=1$

$\sin(x)=1$

$x=\sin^{-1}(1)=90$

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Example Question #1 : Quadratic Formula With Trigonometry

Solve the following equation for $0\leq x<2\pi$ .

$3\sin x=\cos 2x$

Possible Answers:

x=0.285

x=1.286; 4.997

No solution exists

x=0.285; 2.857

Correct answer:

x=0.285; 2.857

Explanation:

$3\sin x=\cos 2x$ ; Use the double angle identity for cosine.

$3\sin x=1-2\sin ^2x$ ; Move everything to the left side of the equation.

$2\sin ^2x+3\sin x-1=0$ ; This is a quadratic-like expression that cannot be factored. We must use the quadratic formula. It may be helpful to see this if you replace $\sin x$ with , so it becomes:

2y^2+3y-1=0

Recall the quadratic formula $y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

plug in a=2; b=3; c=1 .

We now have

$\sin x = \frac{-3\pm \sqrt{17}}{4}$ ; Separate this into two equations and take the inverse sine.

$x = \sin ^{-1}\left( \frac{-3+\sqrt{17}}{4}\right )$ or $x = \sin ^{-1}\left( \frac{-3-\sqrt{17}}{4}\right )$

The first equation gives us x=0.285 . Using the unit circle as we did in previous problems, we can find a second answer from this which is x=2.857 . The second equation will not give us a solution.

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Example Question #1 : Quadratic Formula With Trigonometry

$2\sin^2(x)+5\sin(x)+2=0$

What are the zeros of the function listed above for the interval $[0,2\pi]$ .

Possible Answers:

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

$-\frac{1}{2},-2$

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

$0, 2\pi$

$\pi, 3\pi$

Correct answer:

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

Explanation:

When the quadratic formula is applied to the function, it yields

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-5\pm \sqrt{25-16}}{4}=\frac{5\pm3}{4}= =-\frac{1}{2}, -2$

So those are the zeros for sine, but sine has a minimum of -1, so -2 is out. For -1/2, sine achieves that twice in a cycle, at π+π/6 and 2π-π/6. So while -π/6 is true, it is not correct since it is not in the given interval.

Therefore on the given interval the zeros are:

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

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Example Question #3 : Quadratic Formula With Trigonometry

Solve the following trigonometric equation:

$cos^{2}(x)+cos(x)-2=0$

for $x\in [0,\frac{\pi}{2}]$

Possible Answers:

$\frac{\pi}{2}$

$-\frac{\pi}{2}$

The equation does not have a solution.

Correct answer:

Explanation:

Since $cos^{2}(x)+cos(x)-2=0$ can be written as:

(cosx-1)(cosx+2)=0 . We can't have cosx=-2 .

Therefore cosx-1=0 . This means that $x=2k\pi,$ where k is an integer.

since $\in[0,\p\frac{\pi}{2}]$ . We have x=0 is the only number that satisfies this property.

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Example Question #41 : Solving Trigonometric Equations

Solve each equation over the domain 0 < x < 360 (answer in degrees).

2cos(x) - 1 = 0

Possible Answers:

60, 120

30, 330

60, 300

Correct answer:

60, 300

Explanation:

Rearrange the problem,

2cos(x) - 1 = 0

2cos(x)=1

Over the interval 0 to 360 degrees, cosx = 1/2 at 60 degrees and 300 degrees.

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Trigonometry : Trigonometry

Study concepts, example questions & explanations for Trigonometry

All Trigonometry Resources

Example Questions

Example Question #51 : Trigonometric Equations

Example Question #52 : Trigonometric Equations

Example Question #22 : Finding Trigonometric Roots

Example Question #221 : Trigonometry

Example Question #222 : Trigonometry

Example Question #223 : Trigonometry

Example Question #1 : Quadratic Formula With Trigonometry

Example Question #1 : Quadratic Formula With Trigonometry

Example Question #3 : Quadratic Formula With Trigonometry

Example Question #41 : Solving Trigonometric Equations

All Trigonometry Resources

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