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Popular >

-2sin^2(θ)+sin(θ)+1=0

Solution

- 2 sin^{2} (θ) + sin (θ) + 1 = 0

Solution

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{2} + 2 πn

+1

Solution steps

- 2 sin^{2} (θ) + sin (θ) + 1 = 0

Solve by substitution

sin (θ) = - \frac{1}{2}, sin (θ) = 1

sin (θ) = - \frac{1}{2} : θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

sin (θ) = 1 : θ = \frac{π}{2} + 2 πn

Combine all the solutions

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn, θ = \frac{π}{2} + 2 πn

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7 sin^{2} (x) - 14 sin (x) + 2 = - 5

Frequently Asked Questions (FAQ)

What is the general solution for -2sin^2(θ)+sin(θ)+1=0 ?
The general solution for -2sin^2(θ)+sin(θ)+1=0 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin,θ= pi/2+2pin