What is the general solution for 3sin^2(θ)+6=2sin^2(θ)+7 ?

The general solution for 3sin^2(θ)+6=2sin^2(θ)+7 is θ= pi/2+2pin,θ=(3pi)/2+2pin

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3sin^2(θ)+6=2sin^2(θ)+7

3 sin^{2} (θ) + 6 = 2 sin^{2} (θ) + 7

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

3 sin^{2} (θ) + 6 = 2 sin^{2} (θ) + 7

Solve by substitution

sin (θ) = 1, sin (θ) = - 1

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

10 sin^{2} (4 x) = 5

cosh (x) = 2

sin (2 x) - sin (x) = 0, 0 \leq x \leq 2 π

- 2 cos (3 θ) = 1

sec^{2} (x) + 0.5 tan (x) - 1 = 0

What is the general solution for 3sin^2(θ)+6=2sin^2(θ)+7 ?
The general solution for 3sin^2(θ)+6=2sin^2(θ)+7 is θ= pi/2+2pin,θ=(3pi)/2+2pin