Question

cosxtanx-2cos^2x=-1, solve for all real values

Answer 1

\[\cos x * \tan x = \cos x * \frac{\sin x}{\cos x} = \sin x\]

Answer 2

\[\cos^{2} x = 1-\sin^{2} x\]

Answer 3

?? please help

Answer 4

\[\cos(x)\tan(x)-2\cos^2(x)=-1\]\[\cos(x)\frac{ \sin(x) }{ \cos(x) }-\cos^2(x)=-1\]\[\sin(x)-2\cos^2(x)=-1\]\[\sin(x)-2[\sin^2(x)-1]=-1\]\[\sin(x)-2\sin^2(x)+2=-1\]\[-2\sin^2(x)+\sin(x)=-1-2\]\[\frac{ -2\sin^2(x) }{ -2}+\frac{ \sin(x) }{ -2}=\frac{ 3 }{ 2 }\]\[\sin^2(x)-\frac{ 1 }{ 2}\sin(x)+\frac{ 1 }{ 4 }=\frac{ 3 }{ 2 }+\frac{ 1 }{ 4 }\]\[(\sin(x)-\frac{ 1 }{ 2 })^2=\frac{ 6+1 }{ 4 }\]\[\sqrt{(\sin(x)-\frac{ 1 }{ 4 })^2}=\pm \sqrt{\frac{ 7 }{ 4 }}\]\[\sin(x)-\frac{ 1 }{ 4 }=\pm \frac{ \sqrt{7} }{ 2 }\]\[\sin(x)=\frac{ 1 }{ 4 }\pm \frac{ \sqrt{7} }{ 2 }\]\[x=\sin^{-1} (\frac{ 1 }{ 4 }\pm \frac{ \sqrt{7} }{ 2 })\]\[x=\sin^{-1} (\frac{ 1 }{ 4 }+\frac{ \sqrt{7} }{ 2 })=\sin^{-1} (\frac{ 1+\sqrt{7} }{ 4 })\]\[x=\sin^{-1} (\frac{ 1 }{ 4 }-\frac{ \sqrt{7} }{ 2 })=\sin^{-1} (\frac{ 1-\sqrt{7} }{ 4 })\]