Solve this in interval form, from 0 to 2pi. Round to the nearest hundredth.

$-3sin2\theta=1.5$

Question

Solve this in interval form, from 0 to 2pi. Round to the nearest hundredth.

$-3sin2\theta=1.5$

bloomlocke367 · Answer

@mathmate @pooja195

bloomlocke367 · Answer

@dan815 Will you come help?

bloomlocke367 · Answer

can you help?

bloomlocke367 · Answer

@pooja195

bloomlocke367 · Answer

@ganeshie8 Can you please help?

ganeshie8 · Answer

as a start, divide -3 both sides

ganeshie8 · Answer

$$-3\sin(2\theta) = 1.5 \implies \sin(2\theta) = \frac{1.5}{-3} =-\frac{1}{2} $$

ganeshie8 · Answer

$$\sin(2\theta) = -\frac{1}{2} \implies 2\theta = \sin^{-1}\left(-\frac{1}{2}\right)$$

look up unit circle

bloomlocke367 · Answer

I have the unit circle

bloomlocke367 · Answer

what am I looking for?

ganeshie8 · Answer

see for what angles the sin value is $-\frac{1}{2}$

bloomlocke367 · Answer

sin is the y-value, right?

ganeshie8 · Answer

|dw:1432912910233:dw|

bloomlocke367 · Answer

so, 7pi/6 and 11pi/6

bloomlocke367 · Answer

and I just round that to the nearest hundredth and I'll have my answer?

ganeshie8 · Answer

Yes,
$$2\theta = \frac{7\pi}{6}+2k\pi,~ ~\frac{11\pi}{6}+2k\pi$$

that means

$$\theta = \frac{7\pi}{12}+k\pi,~ ~\frac{11\pi}{12}+k\pi$$

ganeshie8 · Answer

but you want solutions in the interval 0 to 2pi, right ?

bloomlocke367 · Answer

yes

ganeshie8 · Answer

plugin k = 0, 1, 2 etc till you hit 2pi

bloomlocke367 · Answer

why?

ganeshie8 · Answer

$$\theta = \frac{7\pi}{12}+k\pi,~ ~\frac{11\pi}{12}+k\pi$$

plugin $k=0$ and get
$$\theta = \frac{7\pi}{12},~ ~\frac{11\pi}{12}$$

plugin $k=1$ and get
$$\theta = \frac{7\pi}{12}+\pi,~ ~\frac{11\pi}{12}+\pi$$

ganeshie8 · Answer

because any integer value of $k$ gives an angle that is a solution to the given equation

ganeshie8 · Answer

there are infinitely many solutions as there are infinitely many integers
your question is about finding the solutions in the interval 0 to 2pi

bloomlocke367 · Answer

okay.

bloomlocke367 · Answer

and you plug in k=2, and get $\large\frac{7\pi}{12}+2\pi,~\frac{11\pi}{12}+2\pi$

ganeshie8 · Answer

yes clearly thats more than $2\pi$, so we may stop now.

ganeshie8 · Answer

k=0 and k=1 are enough

ganeshie8 · Answer

find the corresponding $\theta$ values
as you can see there are exactly four solutions in the interval 0 to 2pi

bloomlocke367 · Answer

okay, and I just round it?

ganeshie8 · Answer

the question asks you to round it so yeah

bloomlocke367 · Answer

so, 1.83, 2.88, 4.97, and 6.02

ganeshie8 · Answer

Looks good, you may double check with wolfram http://www.wolframalpha.com/input/?i=solve%20-3sin(2%5Ctheta)%3D1.5%2C%200%3C%5Ctheta%3C2pi&t=crmtb01

bloomlocke367 · Answer

Okay, thanks!

ganeshie8 · Answer

yw