Find all solutions to the equation. sinθ=−32\sin\theta=-\frac{\sqrt3}{2}sinθ=−23

θ=4π3+2πn,5π3+2πn\theta=\frac{4\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi nθ=34π+2πn,35π+2πn

Find all solutions to the equation. sin θ = − 3 2 \sin\theta=-\frac{\sqrt3}{2} sin θ = − 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z">

we're asked to find all of the solutions to the equation sine of theta equals negative root 3 over 2. So the first thing we want to do is find the solutions that are on our unit circle already. So So we're looking for all of the angles for which the sine is negative root 3 over 2. Now coming over to my unit circle here, I know that my sine values are going to be negative in quadrants 34. So in quadrant 3, I know that for my angle 4 pi over 3, my sine is going to be negative root 3 over 2. Then in quadrant 4, 5 pi over 3, the sine of this is also negative root 3 over 2, and those are the only two angles for which that is true. So the solutions on my unit circle are 4pi over 3 and 5pi over 3. Now the now that I've found my solutions that are on that unit circle, I can go ahead and just take them and add 2 pi in to each of them in order to get all possible solutions no matter how many rotations I'm going around my unit circle. So these represent my solutions 4pithree+2pithree+2pithree@2pithree. That encompasses all solutions to the equation the sine of beta equals negative root 3 over 2. Thanks for watching and I'll see you in the next one.

Find all solutions to the equation. sin⁡θ=−32\sin\theta=-\frac{\sqrt3}{2}sinθ=−23

θ=4π3+2πn,5π3+2πn\theta=\frac{4\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi nθ=34π+2πn,35π+2πn

θ=π3+2πn,2π3+2πn\theta=\frac{\pi}{3}+2\pi n,\frac{2\pi}{3}+2\pi nθ=3π+2πn,32π+2πn

θ=7π6+2πn,11π6+2πn\theta=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi nθ=67π+2πn,611π+2πn

θ=π6+2πn,5π6+2πn\theta=\frac{\pi}{6}+2\pi n,\frac{5\pi}{6}+2\pi nθ=6π+2πn,65π+2πn

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Linear Trigonometric Equations

Trigonometry

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Linear Trigonometric Equations

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객관식

Find all solutions to the equation.
$\sin\]\theta\)=-\(\frac{\sqrt3}{2}$

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

$$\theta$=$\frac{7\pi}{6}$+2$\pi$ n,$\frac{11\pi}{6}$+2$\pi$ n$

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

$$\theta$=$\frac{\pi}{6}$+2$\pi$ n,$\frac{5\pi}{6}$+2$\pi$ n$

검증된 단계별 안내

Start by understanding that the equation $ \sin\theta = -\frac{\sqrt{3}}{2} $ implies that $ \theta $ is in the third or fourth quadrant, where the sine function is negative.

Recall the reference angle for $ \sin\theta = \frac{\sqrt{3}}{2} $ is $ \frac{\pi}{3} $. This means the angles in the unit circle where sine is $ -\frac{\sqrt{3}}{2} $ are $ \theta = \pi + \frac{\pi}{3} $ and $ \theta = 2\pi - \frac{\pi}{3} $.

Calculate these angles: $ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $ and $ \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} $.

Since the sine function is periodic with period $ 2\pi $, the general solutions for $ \theta $ are $ \theta = \frac{4\pi}{3} + 2\pi n $ and $ \theta = \frac{5\pi}{3} + 2\pi n $, where $ n $ is an integer.

Verify the solutions by substituting back into the original equation to ensure they satisfy $ \sin\theta = -\frac{\sqrt{3}}{2} $.

4:25m

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Trigonometry 때문에 힘들어하고 계신가요?

Find all solutions to the equation. sinθ=−32\(\sin\]\theta\)=-\(\frac{\sqrt3}{2}\)sinθ=−23

다음 시청

Linear Trigonometric Equations 연습 세트

Find all solutions to the equation.
$\sin\]\theta\)=-\(\frac{\sqrt3}{2}$