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

The general solution for sin^2(x)+sin(2x)=1 is x= pi/2+2pin,x=(3pi)/2+2pin,x=0.46364…+pin

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

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.46364 \dots + πn

Solution steps

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

Subtract

1

from both sides

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

Rewrite using trig identities

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

Factor

- cos^{2} (x) + 2 cos (x) sin (x) : cos (x) (- cos (x) + 2 sin (x))

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

Solving each part separately

cos (x) = 0 or - cos (x) + 2 sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

- cos (x) + 2 sin (x) = 0 : x = arctan (\frac{1}{2}) + πn

Combine all the solutions

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

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 0.46364 \dots + πn

T, y = sin (2 T)

sin (4 7^{\circ}) = cos (x)

0.82 = 0.102 cos (3.715 t)

sin^{2} (x) + 0.49 = 1

6 cos^{2} (θ) + sin^{2} (θ) = 4

What is the general solution for sin^2(x)+sin(2x)=1 ?
The general solution for sin^2(x)+sin(2x)=1 is x= pi/2+2pin,x=(3pi)/2+2pin,x=0.46364…+pin