Double root in Trigonometric equation?Consider the function f(x)=sin2x+sin⁡x in the domain [0,2π], find the subset of the domain where f≥0.

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Roy Brady · Accepted Answer

The property that causes this is the boundedness of the sine: there is only one solution per period of sin⁡x=1 and sin⁡x=−1, compared to two for sin⁡x=a,a∈(−1,1).If you plot the graphs y=sin⁡x and y=−1 you will see that the latter line is tangent to the sine curve - i.e. 1+sin⁡x has only double root solutions.

Kingston Lowery · Accepted Answer

Your factorization&amp;nbsp;&amp;nbsp;sinx·1+sinx&amp;nbsp;will also explain how the sign of f(x) behaves. Since&amp;nbsp;(1+sin⁡x)&amp;nbsp;is non-negative for all values of x, the sign of f(x) is entirely determined by the sign of the factor&amp;nbsp;(sin⁡x). As this is negative in quadrants III and IV, f(x) is negative &quot;to either side&quot; of its zero at&amp;nbsp;x=3π2.We can also consider the symmetry of the terms&amp;nbsp;&amp;nbsp;sin2x&amp;nbsp;+&amp;nbsp;sin⁡x&amp;nbsp;&amp;nbsp;.&amp;nbsp;Using the &quot;angle-addition&quot; formula for cosine, we note thatcos⁡(x&amp;nbsp;−&amp;nbsp;3π2)&amp;nbsp;&amp;nbsp;=&amp;nbsp;&amp;nbsp;cos⁡x·cos⁡3π2&amp;nbsp;−&amp;nbsp;sin⁡x·sin⁡3π2=cos⁡x⋅0−sin⁡x⋅(−1)=sin⁡x.Each term in the expression for f(x) has &quot;even symmetry&quot; about the value&amp;nbsp;x=3π2, so their sum also possesses this symmetry. Thus, there is no sign-change for f(x) on &quot;passing through&quot;&amp;nbsp;x=3π2.

Double root in Trigonometric equation? Consider the function

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