How would you solve the following trig equation sin(x)=-0.5 by isolating the x?

Make use of the inverse sine function. Take the inverse sine of both sides of the equation. Of course, within the appropriate limits, the inverse sine of the sine function is the original argument, as is the case with any inverse function: f⁻¹(f(x)) = x.

... sin⁻¹(sin(x)) = sin⁻¹(-0.5)

... x = sin⁻¹(-0.5)

... x = -30°

You need to be careful with inverses of trig functions, because they are only defined over a limited domain and range. The range of the inverse sine function is -90° to 90°, so, for example, sin⁻¹(sin(150°)) = sin⁻¹(0.5) = 30°.

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