![\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}](/tpl/images/0480/7254/fe581.png)
Solve the equation exactly overbthe interval [0,360)
sin^2x-cos^2x=0...
Mathematics, 29.01.2020 06:12 colyernicholas44
Solve the equation exactly overbthe interval [0,360)
sin^2x-cos^2x=0
Answers: 1
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