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xz_007 [3.2K]
3 years ago
14

Solve the equation exactly overbthe interval [0,360) Sin^2x-cos^2x=0

Mathematics
1 answer:
trasher [3.6K]3 years ago
8 0
\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\
-----------------------------\\\\

\bf sin^2(x)-cos^2(x)\implies sin^2(x)-[1-sin^2(x)]=0
\\\\\\
sin^2(x)-1+sin^2(x)=0\implies 2sin^2(x)=1\implies sin^2(x)=\cfrac{1}{2}
\\\\\\
sin(x)=\pm\sqrt{\cfrac{1}{2}}\implies \measuredangle x=sin^{-1}\left( \pm\sqrt{\cfrac{1}{2}} \right)\iff \measuredangle x=sin^{-1}\left( \pm\cfrac{\sqrt{2}}{2} \right)
\\\\\\
\measuredangle x=
\begin{cases}
\frac{\pi }{4}\\\\
\frac{3\pi }{4}\\\\
\frac{5\pi }{4}\\\\
\frac{7\pi }{4}
\end{cases}
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