Question

I'm struggling with this problem : Solve for theta, 0 less than or equal to pi less than or equal to 2pi.

2+cos^2theta = 3sin^2theta

Any help would be greatly appreciated.

Answer 1

\[Solve for \theta : 0 \le \pi \le 2\pi \]
\[equation : 2+\cos ^{2}\theta = 3\sin ^{2}\theta\]

Answer 2

Hint: replace \(\cos^2 \theta\) with \(1-\sin^2 \theta\)

Answer 3

Would I then cross out the \[\sin ^{2}\] ? or is that like completely off base ?

Answer 4

A theta is supposed to be with the sin^2

Answer 5

Why would you do that? Instead you should add \(\sin^2 \theta\) to both sides

Answer 6

If you do that, you'll end up with \(3 = 4\sin^2\theta\)

Answer 7

Then you'll have to divide both sides by 4 and start taking square roots and such

Answer 8

cos^2theta=1-sin^2theta

Answer 9

please have a look at here 2+(1-sin^2x)=3sin^x
2+1-sin^2x=3sin^x
3=3sin^x+sin^2x
3=4sin^x
3/4=sin^x
0.75=sin^x
sinx=0.866
x=60

Answer 10

\[2+\cos ^{2}\theta = 3\sin ^{2}\theta\]\[2+\cos ^{2}\theta +\sin^2\theta= 4\sin ^{2}\theta\]\[3= 4\sin ^{2}\theta\]
\[\arcsin\left(\sqrt\frac{3}{4}\right)=\theta\]

Answer 11

its square so give squaree root on both sides 3/4=sin^x