# Solve: $\left(1-\sin2x\right)\left(\cos x-\sin x\right)=1-2\sin^{2}x$

My attempt:

$$1-2\sin^{2}x=\cos2x$$ Let $$\cos x-\sin x=t$$ Thus, on squaring: $$\sin2x=1-t^{2}$$

I tried simplifying further, but that $$cos2x$$ is giving me trouble.

• What is the original equation you are trying to solve? Mar 4 at 14:37
• i mean it's the title itself
– Vega
Mar 4 at 14:39

After a rewrite, we get $$(\cos x-\sin x)^3 = (\cos x-\sin x)(\cos x+\sin x)$$Can you solve from here?

• I got two solutions, but i'm unable to get the third.
– Vega
Mar 4 at 14:52
• Which one can you not get? Mar 4 at 14:52
• $x=2n\pi+\frac{\pi}{2}$ this one
– Vega
Mar 4 at 14:53
• $\sin\left(2n\pi+\frac\pi2\right)=\cos\left(2n\pi+\frac\pi2\right)$. What does that tell you in the above formula? Mar 4 at 14:55
• Ok thanks, i found my mistake, all clear now!
– Vega
Mar 4 at 15:04

$$\sin2x=2\cos x\sin x \Rightarrow 1-\sin2x=cos^2x+sin^2x-2\sin x \cos x=(cosx-sinx)^2$$

$$1-\sin2x=(cosx-sinx)^2 \Rightarrow (1-\sin2x)(cosx-sinx)=(cosx-sinx)^3=1-2\sin^2x$$

$$(cosx-sinx)^3=1-2\sin^2x=\sin2x$$

$$cosx-sinx=t \Rightarrow t^3=1-t^2$$