What is the general solution for sin(x+0.5)=0.92,0<= x<= 2pi ?

The general solution for sin(x+0.5)=0.92,0<= x<= 2pi is x=0.66808…,x=pi-1.66808…

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sin(x+0.5)=0.92,0<= x<= 2pi

sin (x + 0.5) = 0.92, 0 \leq x \leq 2 π

x = 0.66808 \dots, x = π - 1.66808 \dots

Solution steps

sin (x + 0.5) = 0.92, 0 \leq x \leq 2 π

Apply trig inverse properties

x + 0.5 = arcsin (0.92) + 2 πn, x + 0.5 = π - arcsin (0.92) + 2 πn

Solve

x + 0.5 = arcsin (0.92) + 2 πn : x = arcsin (0.92) + 2 πn - 0.5

Solve

x + 0.5 = π - arcsin (0.92) + 2 πn : x = π - arcsin (0.92) + 2 πn - 0.5

x = arcsin (0.92) + 2 πn - 0.5, x = π - arcsin (0.92) + 2 πn - 0.5

Solutions for the range

0 \leq x \leq 2 π

x = 0.66808 \dots, x = π - 1.66808 \dots

cos (4 x) = 0.8 - 0.6

5 cos^{2} (x) = 4 - 3 sin^{2} (x)

\frac{18 cos ( 2 θ ) + 2 + cos ( θ )}{2} = 0

0 = 3 sin (x - \frac{1}{2} π) + 2

2 sin^{2} (x) = 3 sin (x)

What is the general solution for sin(x+0.5)=0.92,0<= x<= 2pi ?
The general solution for sin(x+0.5)=0.92,0<= x<= 2pi is x=0.66808…,x=pi-1.66808…