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

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Answer 1 · 2016-11-12T20:41:00

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!