Question

Solve the following equations.
$(sin7x+cos7x{)}^2=2si{n}^{2}11x+sin30x.$

Answer 1

$(sin7x+cos7x{)}^2=2si{n}^{2}11x+sin30x$

$\Rightarrow si{n}^{2}7x+co{s}^{2}7x+2sin7x.cos7x=2si{n}^{2}11x+sin30x$

$\Rightarrow 1+sin14x=2si{n}^{2}11x+sin30x$

$\Rightarrow 1-2si{n}^{2}11x=sin30x-sin14x$

$\Rightarrow cos22x=2cos22xsin8x$

$\Rightarrow cos22x(1-2sin8x)=0$

$\therefore cos22x=0$ or $sin8x=1/2$

$\Rightarrow 22x(2n+1)\frac{\pi }{2}\Rightarrow 8x=sin\frac{\pi }{6}$

$\Rightarrow x=(2x+1)\frac{\pi }{44}$ $x=\frac{m\pi }{8}+(-1{)}^m\pi /48$

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