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

The general solution for sin(x)(2sin(x)+1)=2 is x=0.89590…+2pin,x=pi-0.89590…+2pin

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sin(x)(2sin(x)+1)=2

sin (x) (2 sin (x) + 1) = 2

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

Solution steps

sin (x) (2 sin (x) + 1) = 2

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

4 sin (θ) + 1 = 0

sin^{2} (θ) (1 + csc^{2} (θ)) = k

- 3 cos (2 θ) + 9 cos (θ) - 7 = 4 cos (θ) - 3

sin (4 0^{\circ} - x) = cos (3 x)

y, e^{x} - sin (y) = x

What is the general solution for sin(x)(2sin(x)+1)=2 ?
The general solution for sin(x)(2sin(x)+1)=2 is x=0.89590…+2pin,x=pi-0.89590…+2pin