What is the general solution for 2sin^2(x)+1=4sin(x) ?

The general solution for 2sin^2(x)+1=4sin(x) is x=0.29725…+2pin,x=pi-0.29725…+2pin

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2sin^2(x)+1=4sin(x)

2 sin^{2} (x) + 1 = 4 sin (x)

x = 0.29725 \dots + 2 πn, x = π - 0.29725 \dots + 2 πn

Solution steps

2 sin^{2} (x) + 1 = 4 sin (x)

Solve by substitution

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

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

No Solution

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

Combine all the solutions

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

Show solutions in decimal form

x = 0.29725 \dots + 2 πn, x = π - 0.29725 \dots + 2 πn

\frac{1}{4} = cos (x)

cos (2 x) + 3 cos (x) = - 2

4 tan^{2} (θ) + 16 tan (θ) = 7 tan (θ) - 2

10 (1 - cos (θ)) = sin^{2} (θ)

cos (2 x) = 2 - 5 cos (x)

What is the general solution for 2sin^2(x)+1=4sin(x) ?
The general solution for 2sin^2(x)+1=4sin(x) is x=0.29725…+2pin,x=pi-0.29725…+2pin