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Orlov [11]
3 years ago
15

If cos(x)cos(π/7)+sin(x)sin(π/7)= - (√2)/2, then x can equal:

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
8 0

\bf \textit{Sum and Difference Identities} \\\\ sin(\alpha + \beta)=sin(\alpha)cos(\beta) + cos(\alpha)sin(\beta) \\\\ sin(\alpha - \beta)=sin(\alpha)cos(\beta)- cos(\alpha)sin(\beta) \\\\ cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\ cos(\alpha - \beta)= cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta) \\\\[-0.35em] ~\dotfill\\\\

\bf cos(x)cos\left( \cfrac{\pi }{7} \right)+sin(x)sin\left( \cfrac{\pi }{7} \right)=-\cfrac{\sqrt{2}}{2}\implies cos\left( x-\cfrac{\pi }{7} \right)=-\cfrac{\sqrt{2}}{2} \\\\\\ x-\cfrac{\pi }{7}=cos^{-1}\left( -\cfrac{\sqrt{2}}{2} \right)\implies x-\cfrac{\pi }{7}= \begin{cases} \frac{3\pi }{4}\\\\ \frac{5\pi }{4} \end{cases} \\\\[-0.35em] ~\dotfill

\bf x-\cfrac{\pi }{7}=\cfrac{3\pi }{4}\implies x=\cfrac{3\pi }{4}+\cfrac{\pi }{7}~\hfill \stackrel{n ~\in~ \mathbb{Z}}{x=\cfrac{3\pi }{4}+\cfrac{\pi }{7}~~~~+2\pi n} \\\\[-0.35em] ~\dotfill\\\\ x-\cfrac{\pi }{7}=\cfrac{5\pi }{4}\implies x=\cfrac{5\pi }{4}+\cfrac{\pi }{7}~\hfill \stackrel{n ~\in~ \mathbb{Z}}{x=\cfrac{5\pi }{4}+\cfrac{\pi }{7}~~~~+2\pi n}

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