Question

Can anyone help me to solve sinx=cos(x+(π/6)) for -π<x<π?

Answer 1

`x=pi/6`

Explanation:

As `cosx=sin(pi/2-x)`

`cos(x+pi/6)=sin(pi/2-x-pi/6)=sin(pi/3-x)`

As such our equation is `sin(pi/3-x)=sinx`

hence `pi/3-x=x` i.e. `2x=pi/3` and `x=pi/6`

However, other solutions are when

either sum of `pi/3-x` and `x` is `pi` or `0`, which is not the case

Hence only solution is `x=pi/6`