Chủ đề này có 2 trả lời

[ductuMATHER]

Giải PT lượng giác sau: $\frac{sin^{2}x-cos^{2}x+cos^{4}x}{cos^{2}x-sin^{2}x+sin^{4}x}=9$

[NTL2k1]

Giải PT lượng giác sau: $\frac{sin^{2}x-cos^{2}x+cos^{4}x}{cos^{2}x-sin^{2}x+sin^{4}x}=9$ $(*)$

ĐK: $cos^2x-sin^2x+sin^4x\neq 0\Leftrightarrow 1-2sin^2x+sin^4x\neq 0\Leftrightarrow (sin^2x-1)^2\neq 0\Leftrightarrow cos^4x\neq 0\Leftrightarrow cosx\neq 0\Leftrightarrow x\neq \frac{\pi }{2}+k\pi$

$(*)$ $\Leftrightarrow \frac{sin^2x+cos^2x(cos^2x-1)}{cos^2+sin^2x(sin^2x-1)}=9$

$\Leftrightarrow \frac{sin^2x-sin^2cos^2}{cos^2x-sin^2xcos^2x}=9$

$\Leftrightarrow sin^2x-sin^2xcos^2x=9cos^2x-9sin^2xcos^2x$

$\Leftrightarrow 8sin^2xcos^2x=8cos^2x+cos2x$

$\Leftrightarrow 8cos^2x(sin^2x-1)=cos2x\Leftrightarrow -8cos^4x=cos2x$

$\Leftrightarrow -4cos^2x(cos2x+1)=cos2x\Leftrightarrow -2(cos2x+1)^2=cos2x\Leftrightarrow -2cos^2x-5cos2x-2=0$

$\Leftrightarrow$ $\left[ {\begin{array}{*{20}{c}} {cos2x=-\frac{1 }{2}} \\ {cos2x=-2 (loai)} \\ \end{array}} \right.$

$\Rightarrow$ $\left[ {\begin{array}{*{20}{c}} {2x=\frac{2\pi }{3}+k2\pi} \\ {2x=\frac{4\pi }{3}+k2\pi} \\ \end{array}} \right.$

$\Leftrightarrow$ $\left[ {\begin{array}{*{20}{c}} {x=\frac{\pi }{3}+k\pi} \\ {x=\frac{2\pi }{3}+k\pi} \\ \end{array}} \right.$ (thỏa mãn ĐK)

Bài viết đã được chỉnh sửa nội dung bởi NTL2k1: 28-08-2017 - 20:51

[TrollMath]

$(*)$ $\Leftrightarrow \frac{sin^2x+cos^2x(cos^2x-1)}{cos^2+sin^2x(sin^2x-1)}=9$

$\Leftrightarrow \frac{sin^2x-sin^2cos^2}{cos^2x-sin^2xcos^2x}=9$

$\Leftrightarrow sin^2x-sin^2xcos^2x=9cos^2x-9sin^2xcos^2x$

$\Leftrightarrow 8sin^2xcos^2x=8cos^2x+cos2x$

$\Leftrightarrow 8cos^2x(sin^2x-1)=cos2x\Leftrightarrow -8cos^4x=cos2x$

$\Leftrightarrow -4cos^2x(cos2x+1)=cos2x\Leftrightarrow -2(cos2x+1)^2=cos2x\Leftrightarrow -2cos^2x-5cos2x-2=0$

$\Leftrightarrow$ $\left[ {\begin{array}{*{20}{c}} {cos2x=-\frac{1 }{2}} \\ {cos2x=-2 (loai)} \\ \end{array}} \right.$

$\Rightarrow$ $\left[ {\begin{array}{*{20}{c}} {2x=\frac{2\pi }{3}+k2\pi} \\ {2x=\frac{4\pi }{3}+k2\pi} \\ \end{array}} \right.$

$\Leftrightarrow$ $\left[ {\begin{array}{*{20}{c}} {x=\frac{\pi }{3}+k\pi} \\ {x=\frac{2\pi }{3}+k\pi} \\ \end{array}} \right.$

Bạn ko xét điều kiện $cos^{2}x-sin^{2}x+sin^{4}x \neq 0$ à