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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

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    $\begingroup$ Hint: Could you define what "range" means in this context? $\endgroup$ – Mefitico Apr 29 at 16:07
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This is simply asking you the following: For what values of $x$ is $\sin x \geq -\dfrac{1}{2}$? Does that help?

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  • $\begingroup$ Yes it did help a little bit. SinX >= -1/2 how do you find the actual values of X? $\endgroup$ – Sarwin Apr 29 at 15:45
  • $\begingroup$ Plot the curve. For what values in $[0,2\pi]$, $\sin x = -1/2$? Then see for yourself where it is bigger than $-1/2$ $\endgroup$ – Vizag Apr 29 at 16:03
  • $\begingroup$ You are not suppose to draw the actual graph but only sketch it roughly. Then the range should be determined by calculations. $\endgroup$ – Sarwin Apr 29 at 16:17
  • $\begingroup$ Do you know anything about inverse $\sin(x)$, or $\sin^{-1}(x)$? $\endgroup$ – JacobCheverie Apr 29 at 17:01
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The graph sketching part is easy

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.

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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.

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