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

The general solution for 2sin^2(x)+sin(x)-2=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin

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

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

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

Solution steps

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

Solve by substitution

sin (x) = \frac{- 1 + 17}{4}, sin (x) = \frac{- 1 - 17}{4}

sin (x) = \frac{- 1 + 17}{4} : x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

sin (x) = \frac{- 1 - 17}{4} :

No Solution

Combine all the solutions

x = arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = π - arcsin (\frac{- 1 + 17}{4}) + 2 πn

Show solutions in decimal form

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

cos (θ) = \frac{25}{714}

cot^{2} (a) = 1

3 sec^{2} (x) - 4 = 0, 0 \leq x < 2 π

2 sec (θ) + 4 = 0

sin (x) = \frac{( 60 \cdot sin ( 15 5 ^{\circ} ) )}{97.72}

What is the general solution for 2sin^2(x)+sin(x)-2=0 ?
The general solution for 2sin^2(x)+sin(x)-2=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin