Question

Помогите с решением:

Answer 1

$4tg^2x + \frac{3}{cosx} +3 =0;\\ cos x \neq 0;\\ x \neq \frac{\pi}{2}+2\pi k;\\ 4(\frac{1}{cos^2 x}-1)+ \frac{3}{cosx} +3 =0;\\ \frac{4}{cos^2 x}+ \frac{3}{cosx} -4+3 =0;\\ \frac{4}{cos^2 x}+ \frac{3}{cosx} -1 =0;\\ (\frac{4}{cos x}-1)(\frac{1}{cos x}+1)=0;$

откуда cos x=4, что невозможно для любого х: сos x<=1

или $cos x=-1;\\ x=\pi+2 \pi*n$

n єZ

ответ: $\pi+2 \pi*n$

Answer 2

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