x should be theta but I don't where the theta symbol is.
I need to find all the values of x, 0<=x<=2Pi for:
sin^{2}x=3/4
but the sin^{2} has thrown me out, can anyone please help?
I see. The square root of 3/4 is, square root 3/2, which is Pi/3
Therefore as it is Positive in quadrants I and II the angle must be Pi - Pi/3 = 2Pi/3
so for + square root 3/2 the angles are Pi/3 and 2Pi/3
and for - square root 3/2 the angles are 4Pi/3 and 5Pi/3
Yes, good work!
You found the quadrant I solution, then used the identity $\displaystyle \sin(\pi-x)=\sin(x)$ to get the quadrant II solution, then used the identity $\displaystyle -\sin(x)=\sin(-x)$ along with $\displaystyle \sin(x+2\pi)=\sin(x)$ to get the quadrant III and IV solutions.
just think like normal equation when u solve sin,tan and cos the only diffrent is kinda you have to take away or add pi to get to your intervall :P. Took me so long time to understand this and its so simple (I feel stupid that i did not understand this alot early <.<) btw i recomend to always draw unit circle