(toán 11) lượng giác

T

trantien.hocmai

$\text{câu 2} \\
PT \leftrightarrow [1-cos (x-\frac{\pi}{2})].tan ^2x-(1+\cos x)=0 \\
\leftrightarrow (1-\sin x).\sin ^2x-\cos ^2x.(1+\cos x)=0 \\
\leftrightarrow ...$
 
M

mua_sao_bang_98

Câu 2:

$sin^2(\frac{x}{2}-\frac{\pi}{4})tan^2x-cos^2\frac{x}{2}=0$

ĐK: ...

\Leftrightarrow $\frac{1}{2}(1-sinx)\frac{sin^2x}{cos^2x}-cos^2\frac{x}{2}=0$

\Leftrightarrow $(1-sinx)sin^2x-2cos^2x.cos^2\frac{x}{2}=0$

\Leftrightarrow $(1-sinx)sin^2x-cos^2x(1+cosx)=0$

\Leftrightarrow $(1-sinx)(1-cosx)(1+cosx)-(1-sinx)(1+sinx)(1+cosx)=0$
 
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