I need help solving: $\cos(4x)=\sin(2x)$

Question

In need of help solving the equation; $\cos(4x)=\sin(2x)$. I have tried re-writing $\sin(2x)$, but I'm stuck on what to do with $\cos(4x)$

Trick is simple. Pull in big fish with smaller worm on tackle. Dont sub-express $2u$ with $u$ but express $4u$ in terms of $2u$...Somehow getting along the job wont take you anywhere.. $ \sin_{2x}=u ,, \cos (4u)=1- 2 u^2,, 2u^2+u-1=0,,$ factorize $ (2 u -1)(u+1) =0,\quad u= \frac12,-1; u= \pi/6,5 \pi/6... \pi,2\pi $ &c. — Narasimham, Mar 01 '18 at 14:20

score 2 · Answer 1 · edited Jun 12 '20 at 10:38

2

Hint:

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

Substitute this and solve the quadratic in $\sin 2x$.

Alternatively, write $\sin(2x)$ as $\cos(\dfrac{\pi}{2}-2x)$ and then use the general formula for $\cos x = \cos \alpha$ i.e. $x= 2n\pi \pm \alpha$

edited Jun 12 '20 at 10:38

Community

answered Feb 27 '18 at 18:38

Archer

How did you get to cos 4x = 1- 2sin^2 (2x) ? – Michael.W Feb 27 '18 at 18:40
@Michael.W Because $\sin^2a+\cos^2a=1$ – Feb 27 '18 at 18:48
@Michael.W See the proofs of the double angle formulas here – Archer Feb 27 '18 at 18:52
@ Micheal. W Please learn up all the trig double angle formulas – Narasimham Mar 01 '18 at 14:26

score 0 · Answer 2 · answered Feb 27 '18 at 18:41

0

use that $$\cos(x)-\sin(y)=2 \sin \left(-\frac{x}{2}-\frac{y}{2}+\frac{\pi }{4}\right) \sin \left(\frac{x}{2}-\frac{y}{2}+\frac{\pi }{4}\right)$$

answered Feb 27 '18 at 18:41

drhab · Answer 3 · 2018-02-28T12:05:45.533

0

In general: $$\cos\alpha=\cos\beta\iff\alpha=\pm\beta+2n\pi\text{ where }n\in\mathbb Z\tag1$$

Now observe that $$\cos4x=\sin2x=\cos\left(\frac12\pi-2x\right)$$ and apply $(1)$ with $\alpha=4x$ and $\beta=\frac12\pi-2x$.

edited Feb 28 '18 at 12:05

answered Feb 27 '18 at 19:18

drhab