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

The general solution for 4sin^2(x)+1=4sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

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

Solve by substitution

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

- 0.6 = sin (30 t)

1 = tan (\frac{π}{4} + x)

tan (x) = \frac{6.1}{4}

4 cos (θ) = 4 - 4 cos (θ)

sin^{2} (x) + 6 sin (x) + 5 = 0

What is the general solution for 4sin^2(x)+1=4sin(x) ?
The general solution for 4sin^2(x)+1=4sin(x) is x= pi/6+2pin,x=(5pi)/6+2pin