# Solving a trigonometric equation

• Mar 6th 2009, 03:47 PM
Solving a trigonometric equation
5 Cos ^2θ – 2=4 Cos θ
• Mar 6th 2009, 03:53 PM
Jhevon
Quote:

5 Cos ^2θ – 2=4 Cos θ

note that this is the same as $5 \cos^2 \theta - 4 \cos \theta - 2 = 0$.

now, note that this is a quadratic equation in $\cos \theta$. to see this easier, let $y = \cos \theta$, then you have

$5y^2 - 4y - 2 = 0$

now solve for $y$, and then set it equal to $\cos \theta$ to solve for $\theta$
• Mar 6th 2009, 03:54 PM
Prove It
Quote:

5 Cos ^2θ – 2=4 Cos θ

Let $X = \cos{\theta}$.

So the equation becomes

$5X^2 - 2 = 4X$

$5X^2 - 4X - 2 = 0$.

Solving for X using the Quadratic formula gives

$X = \frac{2 + \sqrt{14}}{5}$ and $X = \frac{2 - \sqrt{14}}{5}$.

Since $X = \cos{\theta}$, this gives

$\cos{\theta} = \frac{2 + \sqrt{14}}{5}$ or $\cos{\theta} = \frac{2 - \sqrt{14}}{5}$.

Remember that $-1\leq \cos{\theta} \leq 1$.

Only one of the above equations can satisfy this. Solve for $\theta$ in the appropriate equation.