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Question
Find the general solution of the following equation:
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Solution
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
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