Question

Find the general solution of the following equation:

$$\sin 3x+\cos 2x=0$$

Answer 1

Given,

sin 3x + cos 2x = 0

We know that: sin θ = cos ${\displaystyle (\frac{\pi }{2}-\theta )}$

∴ cos 2x = −sin 3x

⇒ cos 2x = −cos${\displaystyle (\frac{\pi }{2}-3x)}$

We know that: −cos θ = cos (π – θ)

∴ cos 2x = cos${\displaystyle (\pi -(\frac{\pi }{2}-3x))}$

⇒ cos 2x cos ${\displaystyle (\frac{\pi }{2}+3x)}$

If cos x = cos y, implies x = 2nπ ± y, where n ∈ Z. 

From above expression and on comparison with standard equation we have:

${\displaystyle y=(\frac{\pi }{2}+3x)}$

∴ 2x = 2nπ ± ${\displaystyle (\frac{\pi }{2}+3x)}$

Hence, 

${\displaystyle 2x=2n\pi +\frac{\pi }{2}+3x\text{or}2x=2n\pi -\frac{\pi }{2}-3x}$

∴ ${\displaystyle x=-\frac{\pi }{2}-2n\pi \text{or}5x=2n\pi -\frac{\pi }{2}}$

⇒ ${\displaystyle x=-\frac{\pi }{2}(1+4n)\text{or}x=\frac{\pi }{10}(4n-1)}$

∴ ${\displaystyle x=-\frac{\pi }{2}(4n+1)\text{or}\frac{\pi }{10}(4n-1)}$, where n ∈ Z