Question

решите уравнение 6sin²x-5cos x-5=0
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Answer 1

$6\sin^2x-5\cos x-5=0\\6\cdot\left(1-\cos^2x\right)-5\cos x-5=0\\6-6\cos^2x-5\cos x-5=0\\-6\cos^2x-5\cos x+1=0\\6\cos^2x+5\cos x-1=0\\\cos x=t,\;\cos^2x=t^2,\;t\in[-1;\;1]\\\\6t^2+5t-1=0\\D=25-4\cdot6\cdot(-1)=25+24=49\\t_{1,2}=\frac{-5\pm7}{12}\\t_1=-1\\t_2=\frac16\\\\\cos x=-1\Rightarrow x=\pi+2\pi n\\\cos x=\frac16\Rightarrow x=\pm\arccos\left(\frac16\right)+2\pi n,\;n\in\mathbb{Z}$