Question

How do you solve `sinx cos (pi/6) + sin (pi/6) cos x= 1/sqrt2`?

Answer 1

`(pi)/12 and (7pi)/12`

Explanation:

Use the trig identity:
sin a.cos b + sin b.cos a = sin (a + b)
`sin x.cos (pi/6) + sin (pi/6).cos x = sin (x + pi/6).`
The equation transforms to:
`sin (x + pi/6) = 1/sqrt2 = sin (pi/4)`.
The trig unit circle gives another arc `(3pi)/4` that has the same sin value `(1/sqrt2).`
a. `x + pi/6 = pi/4` --> `x = pi/4 - pi/6 = pi/12`
b. `x + pi/6 = (3pi)/4` ---> `x = (3pi)/4 - pi/6 = (7pi)/12`
Answer for `(0, 2pi)`:
`pi/12 and (7pi)/12`