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

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

Find all solutions to the equation. sin θ = − 3 2 \sin\theta=-\frac{\sqrt3}{2} sin θ = − 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 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 = − 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">

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

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

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

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

11. Inverse Trigonometric Functions and Basic Trig Equations

Linear Trigonometric Equations

Precalculus

11. Inverse Trigonometric Functions and Basic Trig Equations

Linear Trigonometric Equations

Struggling with Precalculus?

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Multiple Choice

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

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

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

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

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

Verified step by step guidance

Identify the given equation: $ \sin \theta = -\frac{\sqrt{3}}{2} $. This indicates that we are looking for angles where the sine value is $-\frac{\sqrt{3}}{2}$.

Recall that the sine function is negative in the third and fourth quadrants. The reference angle for $ \sin \theta = \frac{\sqrt{3}}{2} $ is $ \frac{\pi}{3} $.

Determine the angles in the third and fourth quadrants that have the same reference angle. These are $ \theta = \pi + \frac{\pi}{3} = \frac{4\pi}{3} $ and $ \theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3} $.

Express the general solutions for these angles by adding multiples of the full period of the sine function, which is $ 2\pi $. Thus, the solutions 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 checking that they satisfy the original equation $ \sin \theta = -\frac{\sqrt{3}}{2} $ and are consistent with the sine function's periodicity and quadrant rules.

4:25m

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