Question 1206030
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Solve  cos^2(a) + cos(a) = sin^2(a) on the interval 0° ≤ a < 360°.
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<pre>
Replace  sin^2(a) by 1 - cos^2(a).


You will get then

    cos^2(a) + cos(a) = 1-cos^2(a),

or

    2cos^2(a) + cos(a) - 1 = 0.


It is a quadratic equation relative to cos(a), so you can write the solution 
for cos(a) using the quadratic formula

    cos(a) = {{{(-1 +- sqrt(1^2 - 4*2*(-1)))/(2*2)}}} = {{{(-1 +- sqrt(9))/4}}} = {{{(-1 +- 3)/4}}}.


One root is  cos(a) = {{{(-1 + 3)/4}}} = {{{2/4}}} = {{{1/2}}}.

It provides the solutions a = 60° and a = 300° in the given interval.



Other root is cos(a) = {{{(-1-3)/4}}} = -1.

It provides the solution a = 180°.



<U>ANSQER</U>.  The solutions to the given equation are the angles 60°, 180° and 300° in ascending order, in the given interval.
</pre>

Solved.