Question

4cos^2(x) + 4sin(x) - cos(x) - 1 = 0; solve for 0 =< x < 2pi

Answer 1

\[4\cos^2(x) + 4\sin(x) - \cos(x) - 1 = 0; solve for 0 x < 2\pi\]

original question was \[4\cos(x)+4\tan(x)=\sec(x)\]

Answer 2

There's an extra cos in the first line. It should be \(4 \cos^2{x}+4\sin{x}-1=0\).

Answer 3

\[4\cos(x) + 4(\frac{ \sin(x) }{ \cos(x) }) + \frac{ 1 }{ \cos(x) }\]
\[\frac{ 4\cos^2(x) }{ \cos(x) } +\frac{ 4\sin(x) }{ \cos(x) } = \frac{ 1 }{ \cos(x) }\]
\[\frac{ 4\cos^2(x)+4\sin(x)-1 }{ \cos(x) } = 0\]

Answer 4

Remember the pythagorean identity, sin^2(x)+cos^2(x)=1

\[\cos^2(x)+\sin(x)-\frac{ 1 }{ 4 }=0\]
\[1-\sin^2(x)+\sin(x)-\frac{ 1 }{ 4 }=0\]
\[\sin^2(x)-\sin(x)-\frac{ 3 }{ 4 }=0\]

now you can use the quadratic formula or complete the square to get your answer(s) for sin(x) and then use arcsin to find it.