Question

Hello, I need to solve for x.
\[2\sin ^{2}x=3\sin(\frac{ 3\pi }{ 2 }+x)\]
Firstly, do I need to reduction 3\sin(\frac{ 3\pi }{ 2 }+x)\]?

Answer 1

\[2\sin ^{2}x=3\sin(\frac{ 3\pi }{ 2 }+x)\]

Answer 2

@InYourHead

Answer 3

I think you can reduce this first :

\[\sin(\frac{3 \pi}{2} + x) = ??\]

Answer 4

cosx?

Answer 5

How??

Any logic??

Answer 6

I'd love for that to be true, but unfortunately, the truth hurts XD
\[\huge \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\]

Answer 7

cos(x)??

Something left I think, check it further..

Answer 8

Because when there is 3pi/3 fuction chages a "name". So it becomes from sin to cos, right?

Answer 9

@terenzreignz I think there is no need to go by that formula..

Answer 10

and i have forgotten -, right?

Answer 11

correction 3pi/2!

Answer 12

Don't confuse it with
\[\huge \sin\left(\frac{\pi}{2}-\theta\right)=\cos \ \theta\]

Answer 13

is x acute?

Answer 14

Yep,,

So it will be simply -cos(x)

Getting??