Solving a trigonometric equation

**blame_canada100** · Mar 6th 2009, 04:47 PM

5 Cos ^2θ – 2=4 Cos θ

**Jhevon** · Mar 6th 2009, 04:53 PM

Originally Posted by blame_canada100

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 $\theta$

**Prove It** · Mar 6th 2009, 04:54 PM

Originally Posted by blame_canada100

5 Cos ^2θ – 2=4 Cos θ

You have a quadratic equation.

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.

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