Hello mastermin346 Originally Posted by
mastermin346 please help me with this:
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?
Grandad