Question

$sinx = 0 \\ sinx = 1 \\ sinx = - 1 \\ sin 3x = 0 \\ sin \frac{x}{4} = 1 \\ 2in {}^{2} x - 3sinx + 1 = 0 \\ cosx = 1 \\ cosx = - 1 \\ cosx = - \frac{ \sqrt{2} }{2} \\ cosx = \frac{ \sqrt{3} }{2} \\ cos \frac{x}{3} = 0 \\ cos2x = 0 \\ sin(x \frac{\pi}{8} ) = \frac{1}{2} \\ sin {}^{2} 2x = 1 \\ cos( \frac{x}{2} - \frac{\pi}{7} ) = \frac{ \sqrt{3} }{2}$
даю 100 балов !
По алгебре.​

Answer 1

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Объяснение:

На фото..........