Câu 1:
ĐK:$ \cos x \ne 0$
$$\dfrac{\sin^4 x +\cos^4 x}{ \sin2x} = \dfrac{ \tan x + \cos x}{2}$$$$\Longleftrightarrow \dfrac{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}{ 2\sin x\cos x} = \dfrac{ \sin x+\cos^2x}{2}$$$$\Longleftrightarrow \dfrac{1-2\sin^2x\cos^2x}{ 2\sin x\cos x} = \dfrac{ \sin x+\cos^2x}{2}$$$$\Longleftrightarrow 1-2\sin^2x\cos^2x=\sin x(\sin x+\cos^2x)$$$$\Longleftrightarrow \sin^2x+\cos^2x-2\sin^2x\cos^2x=\sin ^2x+\sin x\cos^2x)$$$$\Longleftrightarrow \cos^2x(1-2\sin^2x)=\sin x\cos^2x$$$$\Longleftrightarrow 1-2\sin^2x=\sin x$$$$\Longleftrightarrow 1-2\sin^2x=\sin x$$$$\Longleftrightarrow \left[\begin{array}{1} \sin x=-1(L) \\ \sin x=\dfrac{1}{2} \end{array}\right.$$
Câu 2: $$\tan x = \cot x + 2\cot^3 2x$$
ĐK: $\sin 2x \ne0$
Phương trình tương đương:
$$ 2\cot^3 2x+\cot x-\tan x=0$$$$ \Longleftrightarrow 2\cot^3 2x+\dfrac{\cos^2x-\sin^2x}{\sin x\cos x}=0$$$$ \Longleftrightarrow 2\cot^3 2x +\dfrac{2\cos2x}{\sin 2x}=0$$$$ \Longleftrightarrow \cot^3 2x +\cot 2x=0$$
Câu 3: $$2\sqrt{2}(\sin x + \cos x)\cos x = 3+ \cos2x$$
Phương trình tương đương:
$$ 2\sqrt{2}\sin x\cos x+2\sqrt{2}\cos^2x=3+\cos2x$$$$\Longleftrightarrow \sqrt{2}\sin 2x+\sqrt{2}(1+\cos2x)=3+\cos 2x$$$$\Longleftrightarrow \sqrt{2}\sin 2x+\cos 2x(\sqrt{2}-1)=3-\sqrt{2}$$ Đây là dạng cơ bản rồi nhé.
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
