Question 1110695
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You have three identities for cos(2x):
{{{(cos(x))^2-(sin(x))^2}}}
{{{2(cos(x))^2-1}}}
{{{1-2(sin(x))^2}}}<br>
Since the other term in the equation is sin(x), you want to use the identity for cos(2x) that contains only sin(x); that will give you a quadratic equation in sin(x).  So<br>
{{{cos(2x)-sin(x) = 0}}}
{{{1-2(sin(x))^2-sin(x) = 0}}}
{{{-2(sin(x))^2-sin(x)+1 = 0}}}
{{{2(sin(x))^2+sin(x)-1 = 0}}}
{{{(2(sin(x))-1)(sin(x)+1) = 0}}}
{{{sin(x) = 1/2}}} or {{{sin(x) = -1}}}<br>
The wording of the question is strange: "Determine the exact value of x."  There are an infinite number of values of x that satisfy the equation....<br>
For solutions on the interval [0,2pi), the solutions are {pi/6, 5pi6, 3pi/2}.
A graph:<br>
{{{graph(400,400,0,2pi,-2,2,cos(2x)-sin(x))}}}