1. ## trigonometric problem

solve the following equation for $\displaystyle 0<x<360$.

a)$\displaystyle 4cot^2 x =5 cos x$

b)$\displaystyle 4cos^2 x =9-2 sec^2 x$

2. Hey, when attempting these questions, do not cancel out anything

Here's a start for the first question:

$\displaystyle 4cot^2x = 5cos x$

$\displaystyle 4\frac{\cos^2x}{\sin^2x} - 5\cos x = 0$

$\displaystyle \cos x \left(\frac{4\cos x - 5(1-\cos^2x)}{\sin^2x} \right) = 0$

Solve each term for 0.

3. Originally Posted by Gusbob
Hey, when attempting these questions, do not cancel out anything

Here's a start for the first question:

$\displaystyle 4cot^2x = 5cos x$

still dont get....can give me step by step?

$\displaystyle 4\frac{\cos^2x}{\sin^2x} - 5\cos x = 0$

$\displaystyle \cos x \left(\frac{4\cos x - 5(1-\cos^2x)}{\sin^2x} \right) = 0$

Solve each term for 0.

still dont know..can u teach me step by step sir?

4. Hello mastermin346
Originally Posted by mastermin346

solve the following equation for $\displaystyle 0<x<360$.

a)$\displaystyle 4cot^2 x =5 cos x$
$\displaystyle 4\cot^2 x =5 \cos x$

$\displaystyle \Rightarrow \dfrac{4\cos^2x}{\sin^2x}=5\cos x$
You'll notice that you now have a factor of $\displaystyle \cos x$ on both sides of the equation. So one solution is
$\displaystyle \cos x = 0$
(which you can solve for $\displaystyle 0<\<360^o$)

But if $\displaystyle \cos x \ne 0$ we can divide both sides by $\displaystyle \cos x$ to get:
$\displaystyle \dfrac{4\cos x}{\sin^2x}=5$

$\displaystyle \Rightarrow 4\cos x = 5\sin^2x$
$\displaystyle =5-5\cos^2x$
Now re-arrange this as a quadratic equation in $\displaystyle \cos x$, and solve.

b)$\displaystyle 4cos^2 x =9-2 sec^2 x$
If we put $\displaystyle \cos^2x = c$, the equation can be re-written:
$\displaystyle 4c=9-\dfrac{2}{c}$
Multiply both sides by $\displaystyle c$, and re-arrange this as a quadratic. Solve for $\displaystyle c$ and hence $\displaystyle \cos x$.

Can you complete them both now?