1.
$1+cot2xcotx=\frac{sin2xsinx+cos2xcosx}{sin2xsinx}=\frac{cosx}{sinxsin2x}$
$\Longrightarrow$ $\frac{1+cot2xcotx}{cos^2x}=\frac{1}{sinxcosxsin2x}=\frac{2}{sin^22x}$
Ta lại có đẳng thức quen thuộc : $sin^4x+cos^4x=1-\frac{1}{2}sin^22x$
nên :
$\frac{1+cot2xcotx}{cos^2x}+1=6(sin^4x+cos^4x)$
$\Longleftrightarrow \frac{2}{sin^22x}+1=6-3sin^22x$
$\Longleftrightarrow 3sin^42x-5sin^22x+2=0$
2.
$2cot^2x+1=cot^2x+\frac{1}{sin^2x}=\frac{cos^2x+1}{sin^2x}$
$2tan^2x+1=tan^2x+\frac{1}{cos^2x}=\frac{sin^2x+1}{cos^2x}$
$cos4x=1-sin^22x$
Từ đó :
$\frac{1}{2cot^2x+1}+\frac{1}{2tan^2x+1}=\frac{15cos4x}{8+sin^22x}$
$\Longleftrightarrow$ $ \frac{sin^2x}{1+cos^2x}+\frac{cos^2x}{1+sin^2x}= \frac{15-30sin^22x}{8+ sin^22x}$
$\Longleftrightarrow \frac{1+sin^4x+cos^4x}{(1+cos^2x)(1+sin^2x)}=\frac{15(1-2sin^22x)}{8+sin^22x}$
$\Longleftrightarrow \frac{2-0.5sin^22x}{2+cos^2xsin^2x}=\frac{15(1-2sin^22x)}{8+sin^22x}$
$\Longleftrightarrow \frac{2-0.5sin^22x}{2+0.5sin^22x}=\frac{15(1-2sin^22x)}{8+sin^22x}$
một ẩn rồi nhé
1. $1+cot2xcotx=\frac{sin2xsinx+cos2xcosx}{sin2xsinx}=\frac{cosx}{sinxsin2x}$ $\Longrightarrow$ $\frac{1+cot2xcotx}{cos^2x}=\frac{1}{sinxcosxsin2x}=\frac{2}{sin^22x}$ Ta lại có đẳng thức quen thuộc : $sin^4x+cos^4x=1-\frac{1}{2}sin^22x$ nên : $\frac{1+cot2xcotx}{cos^2x}+1=6(sin^4x+cos^4x)$ $\Longleftrightarrow \frac{2}{sin^22x}+1=6-3sin^22x$ $\Longleftrightarrow 3sin^42x-5sin^22x+2=0$
2. $2cot^2x+1=cot^2x+\frac{1}{sin^2x}=\frac{cos^2x+1}{sin^2x}$ $2tan^2x+1=tan^2x+\frac{1}{cos^2x}=\frac{sin^2x+1}{cos^2x}$ $cos4x=1-sin^22x$ Từ đó : $\frac{1}{2cot^2x+1}+\frac{1}{2tan^2x+1}=\frac{15cos4x}{8+sin^22x}$ $\Longleftrightarrow$ $ \frac{sin^2x}{1+cos^2x}+\frac{cos^2x}{1+sin^2x}= \frac{15-30sin^22x}{8+ sin^22x}$ $\Longleftrightarrow \frac{1+sin^4x+cos^4x}{(1+cos^2x)(1+sin^2x)}=\frac{15(1-2sin^22x)}{8+sin^22x}$ $\Longleftrightarrow \frac{2-0.5sin^22x}{2+cos^2xsin^2x}=\frac{15(1-2sin^22x)}{8+sin^22x}$ $\Longleftrightarrow \frac{2-0.5sin^22x}{2+0.5sin^22x}=\frac{15(1-2sin^22x)}{8+sin^22x}$ một ẩn rồi nhé