Question

solve the following trigonometric equation​

Answer 1

$\mathfrak{Answer}$

Hence, the general solution of sin θ = 1 is θ = (4n + 1)π2, n ∈ Z. Therefore, either, 2 sin x + 3 = 0 ⇒ sin x = - 32, Which is impossible since the numerical value of sin x cannot be greater than 1. We know that the general solution of sin θ = 1 is θ = (4n + 1)π2, n ∈ Z.

Answer 2

$\fbox\orange{A}\fbox\pink{N}\fbox\blue{S}\fbox\red{W}\fbox \purple{E}\fbox\red{R}$

Hence, the general solution of sin θ = 1 is θ = (4n + 1)π2, n ∈ Z. Therefore, either, 2 sin x + 3 = 0 ⇒ sin x = - 32, Which is impossible since the numerical value of sin x cannot be greater than 1. We know that the general solution of sin θ = 1 is θ = (4n + 1)π2, n ∈ Z.