Question 1042505
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Solve sin^2(theta) - 5cos(theta) = 5 for 0° <= theta < 360°.
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Replace {{{sin^2(theta)}}} by {{{1 - cos^2(theta)}}} to have the equation for {{{cos(theta)}}} only. You will get

{{{1-cos^2(theta) - 5cos(theta)}}} = {{{5}}},  or

{{{cos^2(theta) + 5cos(theta) + 4}}} = {{{0}}}.

Factor left side:

{{{(cos(theta)+4)*(cos(theta) +1)}}} = {{{0}}}.

Then the equation deploys in two equations:

1)  {{{cos(theta) + 4}}} = {{{0}}},  or  {{{cos(theta)}}} = {{{-4}}},  which has no solutions,  and

2)  {{{cos(theta) + 1}}} = {{{0}}},  or  {{{cos(theta)}}} = {{{-1}}},  which has the solution {{{theta}}} = 180°.

<U>Answer</U>.  The solution is {{{theta}}} = 180°.
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