1. ## Trig Exam Question

I have had a go at this question but i feel i am on the wrong course as i end up with a far too complicated equation.

This is the question:

Solve, for 0<= θ<180, the equation
4cosec ²2θ º=3+5cot2θ º

Thanks for anyone who can show me how to do this in the most simple way.

P.S. can someone please tell me how to insert the special maths images.

P.P.S. Could do with knowing this by tommorow

2. Originally Posted by Big_Joe
I have had a go at this question but i feel i am on the wrong course as i end up with a far too complicated equation.

This is the question:

Solve, for 0<= θ<180, the equation
4cosec ²2θ º=3+5cot2θ º

Thanks for anyone who can show me how to do this in the most simple way.
$\displaystyle 4\csc^2\left(2\vartheta\right)=3+5\cot\left(2\vart heta\right)$

Recall that $\displaystyle \cot^2\varphi+1=\csc^2\varphi$

Thus, we get $\displaystyle 4\left[\cot^2\left(2\vartheta\right)+1\right]=3+5\cot\left(2\vartheta\right)\implies 4\cot^2\left(2\vartheta\right)-5\cot\left(2\vartheta\right)+1=0$

Making the substitution $\displaystyle z=\cot\left(2\vartheta\right)$ gives us the quadratic equation $\displaystyle 4z^2-5z+1=0$.

This factors to $\displaystyle \left(4z-1\right)\left(z-1\right)=0$.

This produces two different solutions:

$\displaystyle 4z-1=0\implies z=\tfrac{1}{4}$ or $\displaystyle z-1=0\implies z=1$

Since we let $\displaystyle z=\cot\left(2\vartheta\right)$, this now means that:

$\displaystyle \cot\left(2\vartheta\right)=\tfrac{1}{4}$ or $\displaystyle \cot\left(2\vartheta\right)=1$

Can you take it from here?

P.S. can someone please tell me how to insert the special maths images.
See the LaTeX help section here on the forums. You can generate the images by putting the LaTeX code in  tags.

3. Thanks a lot for that, I was on completely the wrong course trying to use double angle formula. I think i can complete this now. If i get the wrong answer i will post back tomorrow.

4. I tried to continue the question but again i feel i am on the wrong track. I tried to find θ by using the double angle formula and saying:

cot2θ=¼
so 1/tan2θ=¼
1/2tanθ/1-tan²θ=¼
1-tan²θ/2tanθ=¼
4(1-tan²θ) = 2tanθ
so 4tan²θ + 2tanθ - 4 = 0
2(2tanθ+2)(tanθ-1)=0
therefore 2tanθ+2 = 0
tanθ = -2/2
θ=tan ‾¹(-1)
θ=-45

and tanθ-1 = 0
θ=tan‾¹(1)
θ=45

Using my method i would also have to do this for the other equation for 2θ that i obtained earlier but the values i have got here are incorrect.
Can someone please tell me where i went wrong!

Edit i have tried to write my calculations using the Latex script:

$\displaystyle \cot{(2\theta)}=\frac{1}{4}$

so$\displaystyle \frac{1}{\tan{(2\theta)}}=\frac{1}{4}$

$\displaystyle \frac{1}{\frac{2\tan\theta}{1-\tan^2\theta}}=\frac{1}{4}$

$\displaystyle \frac{1-\tan^2\theta}{2\tan\theta}=\frac{1}{4}$

$\displaystyle 4(1-\tan^2\theta)=2\tan\theta$

so $\displaystyle 4\tan^2\theta+2\tan\theta-4=0$

$\displaystyle 2(2\tan\theta+2)(\tan\theta-1)=0$

therefore $\displaystyle 2\tan\theta+2=0$

$\displaystyle \tan\theta=\frac{-2}{2}$

$\displaystyle \theta=\arctan{-1}$

$\displaystyle \theta=-45$

and $\displaystyle \tan\theta-1=0$

$\displaystyle \theta=\arctan{1}$

$\displaystyle \theta=45$

5. Originally Posted by Big_Joe
I tried to continue the question but again i feel i am on the wrong track. I tried to find θ by using the double angle formula and saying:

cot2θ=¼
so 1/tan2θ=¼
1/2tanθ/1-tan²θ=¼
1-tan²θ/2tanθ=¼
4(1-tan²θ) = 2tanθ
so 4tan²θ + 2tanθ - 4 = 0
2(2tanθ+2)(tanθ-1)=0
therefore 2tanθ+2 = 0
tanθ = -2/2
θ=tan ‾¹(-1)
θ=-45

and tanθ-1 = 0
θ=tan‾¹(1)
θ=45

Using my method i would also have to do this for the other equation for 2θ that i obtained earlier but the values i have got here are incorrect.
Can someone please tell me where i went wrong!

Edit i have tried to write my calculations using the Latex script:

$\displaystyle \cot{(2\theta)}=\frac{1}{4}$

so$\displaystyle \frac{1}{\tan{(2\theta)}}=\frac{1}{4}$

$\displaystyle \frac{1}{\frac{2\tan\theta}{1-\tan^2\theta}}=\frac{1}{4}$

$\displaystyle \frac{1-\tan^2\theta}{2\tan\theta}=\frac{1}{4}$

$\displaystyle 4(1-\tan^2\theta)=2\tan\theta$

so $\displaystyle 4\tan^2\theta+2\tan\theta-4=0$

$\displaystyle 2(2\tan\theta+2)(\tan\theta-1)=0$

therefore $\displaystyle 2\tan\theta+2=0$

$\displaystyle \tan\theta=\frac{-2}{2}$

$\displaystyle \theta=\arctan{-1}$

$\displaystyle \theta=-45$

and $\displaystyle \tan\theta-1=0$
$\displaystyle \theta=\arctan{1}$

$\displaystyle \theta=45$
It looks alright...but one of your angles doesn't fall in the given interval $\displaystyle 0\leq\vartheta\leq180$!

When you take $\displaystyle \tan^{-1}\left(-1\right)$ you get one angle that is restricted by $\displaystyle 0\leq\vartheta\leq180$. The angle is in the second quadrant. In particular, $\displaystyle \vartheta=135$.

When you take $\displaystyle \tan^{-1}\left(1\right)$ you get one angle that is restricted by $\displaystyle 0\leq\vartheta\leq180$. The angle is in the first quadrant. In particular, $\displaystyle \vartheta=45$

So your solution set would be $\displaystyle \color{red}\boxed{\vartheta=45,~135}$

Does this make sense?

6. Yes but the actual answers to the question are 22.5,112.5,38 and 128. Something must be wrong.

7. Originally Posted by Big_Joe
Yes but the actual answers to the question are 22.5,112.5,38 and 128. Something must be wrong.
Oh..

They went ahead and directly evaluate it...

$\displaystyle \cot\left(2\vartheta\right)=1\implies \vartheta=\tfrac{1}{2}\cot^{-1}\left(1\right)$

But $\displaystyle \cot^{-1}\left(1\right)=45$ or $\displaystyle \cot^{-1}\left(1\right)=225$

Thus, $\displaystyle \vartheta=\tfrac{1}{2}\left(45\right)=\color{red}\ boxed{22.5}$ or $\displaystyle \vartheta=\tfrac{1}{2}\left(225\right)=\color{red} \boxed{112.5}$

But you will have to approximate the other two using a calculator...

$\displaystyle \cot\left(2\vartheta\right)=\tfrac{1}{4}\implies \vartheta=\tfrac{1}{2}\cot^{-1}\left(\tfrac{1}{4}\right)$

But $\displaystyle \cot^{-1}\left(\tfrac{1}{4}\right)\approx 76$ or $\displaystyle \cot^{-1}\left(\tfrac{1}{4}\right)\approx 256$

Therefore, $\displaystyle \vartheta=\tfrac{1}{2}\left(76\right)=\color{red}\ boxed{38}$ or $\displaystyle \vartheta=\tfrac{1}{2}\left(256\right)=\color{red} \boxed{128}$

So our solution set is $\displaystyle \color{red}\boxed{\vartheta=22.5,~38,~112.5,~128}$