Question

Solve the equation `sin(2x) +3cos(2x) = 0` ,for `0<x<180`?

Answer 1

`x ~~ = 54.2^@` and `x ~~ 141.2^@`

Explanation:

Given: `sin(2x) +3cos(2x) = 0`

Subtract `3cos(2x)` from both sides:

`sin(2x) = -3cos(2x)`

Divide both sides by `cos(2x)`

`sin(2x)/cos(2x) = -3; 0^@ < x < 180^@`

Substitute `tan(2x)` for `sin(2x)/cos(2x)`:

`tan(2x) = -3; 0^@ < x < 180^@`

Use the inverse tangent function on both sides:

`2x = tan^-1(-3); 0^@ < x < 180^@`

Divide both sides by 2:

`x = 1/2tan^-1(-3)`

If we were to use a calculator to evaluate the above equation, we would obtain a negative angle that would be outside of the specified domain.

We need to add the cyclical nature of the inverse tangent function that has a period of `180^@`:

`x = 1/2{n(180^@)+tan^-1(-3)}; n in ZZ`

To obtain the two values within the domain, we evaluate the above equation with `n = 1` and `n = 2`:

`x = 90^@+ 1/2tan^-1(-3); 0^@ < x < 180^@`

`x ~~ = 54.2^@`

`x = 180^@ + 1/2tan^-1(-3); 0^@ < x < 180^@`

`x ~~ 141.2^@`