trigonometry problem

**ngkh87** · December 23rd 2009, 10:43 PM

Find x of the equation sin5x+sin3x=cosx

**Lucio Carvalho** · December 24th 2009, 12:46 AM

Hello ngkh87,
Here it is my point of view about the exercise.
I hope you will understand!
Bye!

**Soroban** · December 24th 2009, 05:47 AM

Hello, ngkh87!

My solution is similar to Lucio's . . .

Solve for $x\!:\;\;\sin5x+\sin3x\:=\:\cos x$

We have the Sum-to-Product Identity: . $\sin A + \sin B \:=\:2\sin\left(\frac{A+B}{2}\right)\cos\left(\fra c{A-B}{2}\right)$
. . which Lucio derived for us.

The equation becomes: . $2\sin\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5 x-3x}{2}\right) \:=\:\cos x$

And we have: . $2\sin4x\cos x \:=\:\cos x \quad\Rightarrow\quad 2\sin4x\cos x - \cos x \:=\:0$

Factor: . $\cos x(2\sin4x-1) \:=\:0$

We have two equations to solve:

. . $\cos x \:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:\tfrac{\pi}{2} + \tfrac{\pi}{2}n}$

. . $2\sin4x-1\:=\:0 \quad\Rightarrow\quad \sin4x \:=\:\tfrac{1}{2} \quad\Rightarrow\quad 4x \:=\:\begin{Bmatrix}\frac{\pi}{6} + 2\pi n \\ \\[-4mm] \frac{5\pi}{6} + 2\pi n \end{Bmatrix}$
. . Hence: . $\boxed{x \;=\;\begin{Bmatrix}\frac{\pi}{24} + \frac{\pi}{2}n \\ \\[-3mm] \frac{5\pi}{24} + \frac{\pi}{2}n \end{Bmatrix}}$

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