Solve sin 3x + cos 2x = 0

sin 3x + cos 2x = 0

Let us simplify,

sin 3x + cos 2x = 0

cos 2x = – sin 3x

cos 2x = – cos (π/2 – 3x) [since, sin A = cos (π/2 – A)]

cos 2x = cos (π – (π/2 – 3x)) [since, -cos A = cos (π – A)]

cos 2x = cos (π/2 + 3x)

2x = 2nπ ± (π/2 + 3x)

So,

2x = 2nπ + (π/2 + 3x) [or] 2x = 2nπ – (π/2 + 3x)

x = -π/2 – 2nπ [or] 5x = 2nπ – π/2

x = -π/2 (1 + 4n) [or] x = π/10 (4n – 1)

x = – π/2 (4n + 1) [or] π/10 (4n – 1)

∴ the general solution is

x = – π/2 (4n + 1) [or] π/10 (4n – 1)

x = π/2 (4n – 1) [or] π/10 (4n – 1), where n ϵ Z.

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