Question 826309
{{{cos(2theta)+8cos(theta)+9=0}}}
To rewrite this in terms of {{{cos(theta)}}} we will need to replace the {{{cos(2theta)}}}. There are three variations of this formula:<ul><li>{{{cos(2theta) = cos^2(theta)-sin^2(theta)}}}</li><li>{{{cos(2theta) = 2cos^2(theta)-1)}}}</li><li>{{{cos(2theta) = 1-2sin^2(theta)}}}</li></ul>Any of them will do but since we only want cos, not sin, we will use the one in the middle:
{{{(2cos^2(theta)-1)+8cos(theta)+9=0}}}
Combining the -1 and +9:
{{{2cos^2(theta)+8cos(theta)+8=0}}}
We can make this simpler if we divide both sides by 2 (since 2 is a factor of every term):
{{{cos^2(theta)+4cos(theta)+4=0}}}<br>
The equation we now have is a quadratic equation for {{{cos(theta)}}}. So we can use the discriminant, {{{b^2-4ac}}}, to terminate the number and types of roots for {{{cos(theta)}}}. With a "a" of 1, a "b" of 4 and a "c" of 4 we get:
{{{b^2-4ac = (4)^2-4(1)(4) = 16 - 4(1)(4) = 16 - 16 = 0}}}
A discriminant of 0 means: two equal real roots.<br>
To determine what the roots for {{{theta}}} are, we need to solve the equation. The quadratic equation fits the {{{a^2+2ab+b^2 = (a+b)^2}}} pattern with "a" equal to {{{cos(theta)}}} and "b" equal to 2. So the equation factors according to the pattern:
{{{(cos(theta)+2)^2 = 0}}}
From the Zero Product Property:
{{{cos(theta)+2 = 0}}}
Subtracting 2 we get:
{{{cos(theta) = -2}}}
Here we run into a problem. <i>All</i> cos's have values between -1 and 1. It is not possible for cos of anything to be a -2. So there is no solution for your equation.