Thread: Solving a trigonometric equation

5 Cos ^2θ – 2=4 Cos θ

Originally Posted by blame_canada100

5 Cos ^2θ – 2=4 Cos θ

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

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

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

now solve for $\displaystyle y$, and then set it equal to $\displaystyle \cos \theta$ to solve for $\displaystyle \theta$

Originally Posted by blame_canada100

5 Cos ^2θ – 2=4 Cos θ

You have a quadratic equation.

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

So the equation becomes

$\displaystyle 5X^2 - 2 = 4X$

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

Solving for X using the Quadratic formula gives

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

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

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

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

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