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 `(π/2 - theta)`

∴ cos 2x = −sin 3x

⇒ cos 2x = −cos`(pi/2- 3x)`

We know that: −cos θ = cos (π – θ)

∴ cos 2x = cos`(pi - (pi/2 - 3x))`

⇒ cos 2x cos `(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:

`y = (pi/2 + 3x)`

∴ 2x = 2nπ ± `(pi/2 + 3x)`

Hence, 

`2x = 2npi + pi/2 + 3x or 2x = 2npi - pi/2 - 3x`

∴ `x = -pi/2 - 2npi or 5x = 2npi - pi/2`

⇒ `x = -pi/2 (1 + 4n) or x = pi/10 (4n - 1)`

∴ `x = -pi/2 (4n + 1) or pi/10 (4n - 1)`, where n ∈ Z