4 cos^2 x =2+2sinx .. wht is the solution of the equation in the interval [0,2pie) ?

i got no solution but i m nt sure

Question

4 cos^2 x =2+2sinx .. wht is the solution of the equation in the interval [0,2pie) ? 

i got no solution but i m nt sure

anonymous · Answer

$$

4 \cos^2 (x) =2+2\sin (x)\
4( 1- \sin^2(x))= 2+2\sin (x)\

4 -4\sin^2(x)= 2+2\sin (x)=0\
-4 +4\sin^2(x)+ 2+2\sin (x)=0\
4\sin^2(x)+2\sin (x)-2=0\
2\sin^2(x)+\sin (x)-1=0\
$$
Can you  finish it now?

anonymous · Answer

Use the quadratic formula to find

$$
\sin(x) = -1\
\sin(x) = \frac 1 2
$$

anonymous · Answer

$$
x= \frac {3 \pi}2\
x= \frac \pi 6\
x= \frac {5\pi}6
$$

anonymous · Answer

thanks a lot

anonymous · Answer

hey eliasaab if u could just check my answers for this question it would be a great help

anonymous · Answer

the sign of the tan should be negative
the sign of the sec should be positive

anonymous · Answer

You can practice trig problems on my site http://www.math.missouri.edu/escgi/mucgi-bin/munew.cgi?variable=trig