giải phương trình lượng giác

T

trantien.hocmai

$$2\cos 2x-8\cos x+7=\frac{1}{\cos x} \text{ }(\cos x \not= 0) \\
\leftrightarrow 4\cos ^2x-8cos x+5=\frac{1}{\cos x} \leftrightarrow 4\cos ^3x-8\cos ^2x+5\cos x-1=0 \leftrightarrow ...\\
(1-\tan x)(1+\sin x)=1+\tan x \text{ } \cos x \not= 0 \\
\leftrightarrow (\cos x-\sin x)(1+\sin x)=\sin x+\cos x \leftrightarrow \sin ^2x-\sin x.\cos x+2\sin x=0 \leftrightarrow ... \\
8\cos ^3(x+\frac{\pi}{3})=\cos 3x \\
\text{đặt t=} x+\frac{\pi}{3} \rightarrow x=t-\frac{\pi}{3} \\
8\cos ^3t=\cos (3t-\pi) \leftrightarrow 8\cos ^3t=-\cos 3t \leftrightarrow 12\cos ^3t-3\cos t=0 \\$$
 
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