Question

How do you solve `2\sin x - \cos x = \frac { 3} { \sqrt { 2} }`?

Answer 1

`x=180^@xxn+(-1)^n71.565^@+26.565^@`, where `n` is an integer.

Explanation:

`2sinx-cosx=3/sqrt2`

`hArrsqrt5(sinx xx2/sqrt5-cosx xx 1/sqrt5)=3/sqrt2`

Let `sinalpha=1/sqrt5` where `alpha` is in `Q1`, observe that then `cosalpha=2/sqrt5` and `tanalpha=1/2` and `alpha=tan^(-1)(1/2)=tan26.565^@`

and our equation becomes

`sin(x-alpha)=3/sqrt10=0.9486833=sin71.565^@`

i.e. `x-26.565^@=180^@xxn+(-1)^n71.565^@`, where `n` is an integer.

i.e. `x=180^@xxn+(-1)^n71.565^@+26.565^@`

and in `[0,2pi)`, `x=98.13^@` or `135^@` (for `n=0" and "1`.