Question

Question #40645

Answer 1

`x=(7pi)/6+2kpi,(11pi)/6+2kpi`, where `k` is an integer

Explanation:

We have the equation:

`15sinx+7=sinx`

Subtract `7` from both sides.

`15sinx=sinx-7`

Subtract `sinx` from both sides.

`14sinx=-7`

Divide both sides by `14`.

`sinx=-1/2`

If we are restricting our domain from `0<=x<2pi`, our solutions are at the reference angles of `pi/6` in `"QIII"` and `"QIV"`, when `sinx` is negative. These values give us

`x=(7pi)/6,(11pi)/6`

However, if we don't restrict our domain, these two values and any angle `2pi` away will also satisfy the equation, thus the full solution is:

`x=(7pi)/6+2kpi,(11pi)/6+2kpi`, where `k` is an integer