Question 1100483
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{{{sqrt(3)*sin(2x) + 2 (cosx)^2 = -1}}}<br>
Clearly we need to get trig functions of a single angle, so let's use the double angle identity for sin(2x):<br>
{{{sqrt(3)*2(sinx)(cosx) + 2 (cosx)^2 = -1}}}
{{{sqrt(3)*2(sinx)(cosx) + 2 (cosx)^2 + 1 = 0}}}<br>
Our equation has a mixture of sinx and cosx to various powers; but it also has that "+1" that makes it impossible to work with the equation in this form.<br>
But here is a place where you use the basic trig identity {{{(sinx)^2+(cosx)^2 = 1}}} in an unusual way to make the problem easy to solve.<br>
{{{sqrt(3)*2(sinx)(cosx) + 2 (cosx)^2 + (sinx)^2 + (cosx)^2 = 0}}}
{{{3*(cosx)^2 + 2*sqrt(3)(sinx)(cosx) + (sinx)^2 = 0}}}
{{{(sqrt(3)*cosx + sinx)^2 = 0}}}
{{{sqrt(3)*cosx + sinx = 0}}}
{{{sqrt(3)*cosx = -sinx}}}
{{{sqrt(3) = -sinx/cosx = -tanx}}}
{{{tanx = -sqrt(3)}}}<br>
We need a reference angle of pi/3 in the 2nd and 4th quadrants; the solutions between 0 and 2pi are 2pi/3 and 5pi/3.