Question

Find all real numbers: 3sin^2x-sinx=0 HELP! I just basically need help knowing when i add/sub pi or 2pi

Answer 1

6sin^2 (x) cos^2 (x) -sinx=0
Take sinx common you get sin(x)=0 As you know sin(n pi) =0 so x= n*pi where n is an integer... For remaining portion 6sinx cos^2 (x)-1=0
we get
sinx cos^2 x=1/6
as cos^2 x= 1-sin^2 x
sinx -sin^3 x=1/6 solve the cubic equation

Answer 2

why not just factor out sin(x)

\[\sin(x)[3\sin(x) -1] = 0\]

so fins the values where

sin(x) = 0 and sin(x) = 1/3

they are the solutions.

Answer 3

i lknow you take the inverse sin for both but then it tells me to sub tract the inverse of 1/3 from pi....and idk why

Answer 4

will if you know the sin curve it looks like



so that covers sin(x) = 0

now for the other angle

\[x = \sin^{-1}(\frac{1}{3})\]

just type it into a calculator.