Question

solve for x:

Answer 1

x = 4

Answer 2

\[2 \sin ^2 x - coz x -1 = 0 for 0 \le x < 2\pi. \]

Answer 3

gimmick is to rewrite
\[\sin^2(x)=1-\cos^2(x)\] and then get a quadratic in cosine

Answer 4

\[2(1-\cos^2(x))-\cos(x)-1=0\]
\[2-2\cos^2(x)-\cos(x)-1=0\]
\[2\cos^2(x)+\cos(x)-1=0\] and then solve as if it was
\[2z^2+z-1=0\]

Answer 5

thanks again for your input satellite73! the question states that the answer should be in terms of pi. does this change the calculation?

Answer 6

no. just solve for z, and substitute Cos(x) in when you get your answer. Take inverse cosine and you will get your X term in terms of pi.

Answer 7

thanks mathandphysics! you guys are life-savers! much appreciated.