# Trigonometric Curves, Finding Range

Sketch the graph of $$y=2\sin x + 1$$ for intervals $$0° \leq x \leq 360°$$. Hence state the range of values of $$x$$ in this interval which satisfies the inequality $$2\sin x + 1 \geq 0$$.

The graph sketching part is easy but please can anyone explain how to find the range. Thanks

• Hint: Could you define what "range" means in this context? – Mefitico Apr 29 at 16:07

This is simply asking you the following: For what values of $$x$$ is $$\sin x \geq -\dfrac{1}{2}$$? Does that help?
• Plot the curve. For what values in $[0,2\pi]$, $\sin x = -1/2$? Then see for yourself where it is bigger than $-1/2$ – Vizag Apr 29 at 16:03
• Do you know anything about inverse $\sin(x)$, or $\sin^{-1}(x)$? – JacobCheverie Apr 29 at 17:01
If you have sketched the graph y=2sin x+1, then the question requires the values for which $$y\ge 0$$ i.e.,the regions of graph above x-axis.
If you solve the equation $$2\sin x+1=0$$ in the specified range, you get $$x=210°,330°.$$ You can check that $$2\sin x+1$$ becomes $$1$$ at both endpoints of the given interval; thus by continuity it follows that the quantity is negative only for $$x$$ strictly between $$210°$$ and $$330°.$$ This gives you what you want.