Question

What are two different cases when a trigonometric equation would have no solution?

Answer 1

There are so many answers possible, but here is one such answer.

a) Solve `sin^2x + 7sinx + 10 = 0`.

Letting `t= sinx`

`:.t^2 + 7t + 10 = 0`

`(t + 5)(t + 2) = 0`

`t = -5 and t = -2`

`sinx= -5 and sinx= -2`

`x = arcsin(-5) and x = arcsin(-2)`

But since `y= arcsinx` has a domain of `[-1, 1]`, this equation has no solution.

b) Solve `-csc(x - 2) - 1/2 = 0`

Start by isolating the `csc(x- 2)`.

`-csc(x- 2) = 1/2`

`csc(x - 2) = -1/2`

`1/(sin(x- 2)) = -1/2`

`2 = -sin(x- 2)`

`-2 = sin(x - 2)`

`arcsin(-2) = x- 2`

`x = O/`

Once again, it comes to the limited domain of the sine/cosine function, and hence by extension, the limited domain of the cosecant function.

The rule of thumb with trigonometric equations is that there will be no solutions whenever `-1 > sinx`, `-1>cosx`, `1< sinx` and `1 < cosx`.

Hopefully this helps!