What is the general solution for 2sin^2(x)+2sin(x)-1=0 ?

The general solution for 2sin^2(x)+2sin(x)-1=0 is x=0.37473…+2pin,x=pi-0.37473…+2pin

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2sin^2(x)+2sin(x)-1=0

2 sin^{2} (x) + 2 sin (x) - 1 = 0

x = 0.37473 \dots + 2 πn, x = π - 0.37473 \dots + 2 πn

Solution steps

2 sin^{2} (x) + 2 sin (x) - 1 = 0

Solve by substitution

sin (x) = \frac{- 1 + 3}{2}, sin (x) = - \frac{1 + 3}{2}

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

sin (x) = - \frac{1 + 3}{2} :

No Solution

Combine all the solutions

x = arcsin (\frac{- 1 + 3}{2}) + 2 πn, x = π - arcsin (\frac{- 1 + 3}{2}) + 2 πn

Show solutions in decimal form

x = 0.37473 \dots + 2 πn, x = π - 0.37473 \dots + 2 πn

csc (x) = 7

2 cos (θ) sin (θ) + cos (θ) - 2 sin (θ) = 1

sin (2 θ) - cos (θ) - 2 sin (θ) + 1 = 0

2 cos (x) + 1 = 0 - 2, π \leq x \leq 2 π

cos^{3} (x) = - cos (x)

What is the general solution for 2sin^2(x)+2sin(x)-1=0 ?
The general solution for 2sin^2(x)+2sin(x)-1=0 is x=0.37473…+2pin,x=pi-0.37473…+2pin