Question

Please help me solve this

Answer 1

Answer:
`(\pi)/(3)`

Explanation:

`sin(A+B)=sinAcosB+cosAsinB` and
`sin(A-B)=sinAcosB-cosAsinB`

Applying these two rules into the equation, we can get a new equation

`sin(x)cos((\pi)/(6))+cos(x)sin((\pi)/(6)) - [sin(x)cos((\pi)/(6))-cos(x)sin((\pi)/(6))] = (1)/(2)`

We know
`cos((\pi)/(6) ) = (√(3) )/(2)` and
`sin((\pi)/(6) ) = (1)/(2)` so input this into the equation.

`sin(x)((√(3))/(2))+cos(x)((1)/(2)) - sin(x)((√(3))/(2))+cos(x)((1)/(2)) = (1)/(2)`

We can cancel out the terms
`sin(x)((√(3))/(2) )` and 2 of
`cos(x)((1)/(2))` make
`cos(x)`, so all this make

`cos(x) = (1)/(2)`

So x is equal to
`(\pi)/(3)` in a fixed interval