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

Degrees

x = 38.27819 \dots^{\circ}, x = 84.42602 \dots^{\circ}

Solution steps

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

Apply trig inverse properties

sin (x + 0.5) = 0.92

General solutions for

sin (x + 0.5) = 0.92

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

x + 0.5 = arcsin (0.92) + 2 πn

Move

0.5

to the right side

x + 0.5 = arcsin (0.92) + 2 πn

Subtract

0.5

from both sides

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

Simplify

x = arcsin (0.92) + 2 πn - 0.5

x = arcsin (0.92) + 2 πn - 0.5