Question

How do you solve `sin(x+60°) + cos(x+30°) = 1/2` in the range 0° < x < 360°?

Answer 1

`x = 73.22^@` and `x = 360 - 73.22 = 286.78^@`

Explanation:

Use the trig identities:
`sin (a + b) = sin a.cos b + sin b.cos a`
`cos (a + b) = cos a.cos b - sin a.sin b`

`sin (x + 60) = (1/2).sin x + ((sqrt3)/2).cos x`

`cos (x + 30) = cos x.((sqrt3)/2) - (1/2).sin x`

Adding the 2 expressions, we get:

`(sqrt3).cos x= 1/2 -> cos x = 1/(2sqrt3) = (sqrt3)/6`

->`x = 73.22^@` and `x = 360 - 73.22 = 286.78^@`