Question 1043276
.
Solve the equation for exact solutions over the interval [0, 2π).

cos2x + 2 cos x + 1 = 0
~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
{{{cos^2(x) + 2cos(x) + 1}}} = {{{0}}},  --->  ( apply the formula {{{a^2 + 2a + 1}}} = {{{(a+1)^2}}} )  --->

{{{(cos(x)+1)^2}}} = {{{0}}},  --->

cos(x) + 1 = 0  --->

cos(x) = -1  --->

x = {{{pi}}}.

<U>Answer</U>.  x = {{{pi}}}.
</pre>

Solved.


Or, if your original equation is {{{cos(2x) + cos(x) + 1}}} = {{{0}}}, 

then see the solution under this link

<A HREF=https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.906637.html>https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.906637.html</A>


https://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.906637.html