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Solve the equation :

$$2\sin 2x-2\cos2x=\frac{\cos x+\cos3x}{\cos x-\sin x}$$

my try :

$$\cos x-\sin x \neq 0 \to \cos x \neq \sin x$$

$$x\neq k\pi+\frac{\pi}{4}$$

$$2(\sin 2x -\cos 2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$

$$2(2\sin x \cos x -\cos 2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$

$$2(2\sin x \cos x -(1-2\sin^2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$

then ?

  • 0
    Simplify $\cos x-\cos3x$ – 2017-01-24

1 Answers 1

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Hint:

$\cos3x+\cos x=2\cos2x\cos x$

$\sin2x-\cos2x=\cos x(\cos x+\sin x)$

$\iff \cos^2x+\cos2x=\sin x\cos x$

Divide both sides by $\cos^2x$