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andrezito [222]
3 years ago
7

Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx

Mathematics
1 answer:
Alisiya [41]3 years ago
3 0
\bf cos^4(x)=cos^4(x)csc(x)\\\\
-----------------------------\\\\
cos^4(x)=cos^4(x)\cfrac{1}{sin(x)}\implies cos^4(x)=\cfrac{cos^4(x)}{sin(x)}
\\\\\\
cos^4(x)sin(x)=cos^4(x)\implies cos^4(x)sin(x)-cos^4(x)=0
\\\\\\
cos^4(x)[sin(x)-1]=0\to 
\begin{cases}
cos^4(x)=0\to x=cos^{-1}(0)\\
----------\\
sin(x)-1=0\\
sin(x)=1\to x=sin^{-1}(1)
\end{cases}

\bf \measuredangle x = cos^{-1}(0)\implies \measuredangle x = 
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}
\\\\\\
\measuredangle x=sin^{-1}(1)\implies \measuredangle x=\frac{\pi }{2}

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