Trigonometry Equations - HELP!

I'm currently doing C2 in A Level Maths atm but I've seen that the particular question I am struggling on is in C3?? So can someone please help me.

1) Show that the equation

tan 2x = 5 sinx 2x

can be written in the form

(1-5 cos 2x) sin 2x = 0

Here I just rearranged the first equation using: sin 2x / cos 2x = 5 sin 2x

2) Hence solve for 0 <= x <= 180

tan 2x = 5 sin 2x

Answer to 1dp where appropriate. You must show clearly how answers are obtained


I have attempted this in too many ways and I am just confusing myself. Please Help

Could you post some working so we can help? The Maths forum does not encourage full solutions to be posted.

Here I just rearranged the first equation using: sin 2x / cos 2x = 5 sin 2x

Now multiply both sides by $\cos 2x$ and bring everything over to one side and factorise.

I have done that already, but I am struggling with solving

So i done sin2x/cos 2x = 5sin 2x

sin 2x = 5sin2x x cos 2x

sin 2x - 5 sin 2x x cos 2x = 0
(1-5cos2x) sin x = 0

And then would cos x = 1/5???

What happens when you have a quadratic $(x-2)(x-3) = 0$? You get two solutions, right? By equating each bracket to 0?

In thise case, you have two solution sets as well, one is generated by $1 - 5\cos 2x = 0 \Rightarrow \cos 2x = \frac{1}{5}$ (solve this and find all the solutions)

The other is given by $\sin x = 0$ (solve this and find all the solutions)

Then put all these solutions in one list x = 0, pi, etc... from both of the above solutions and that's your answer.

Okay, thank you

I shall try it and then post my answers to double check

Awesome, will check them for you.

Okay because it is for 0<= x <= 180

I got for cos x

39.3 (1dp) and 140.8 (1dp)

And then sin x

0 and 180

Fingers crossed it's correct

These are both correct.



Sorry! I wrote $\sin x = 0$ when I should have written $\sin 2x = 0$

So would sin x just be 0 and 90?

Almost! $\sin 2x = 0 \Rightarrow 2x = 0, 180^{\circ}, 360^{\circ}$, so $x = 0, 90^{\circ}, ?$.

Okay so I would get

Sin x = 0, 90 and 180??

You keep writing sin x = 0, 90, 180. But you need to write x = 0, 90, 180! (*) Otherwise, those values are correct, yes.


(*) It's like me saying solve $x^2 = 9$ and saying the solutions are $x^2 = 3, -3$. That's obviously nonsense, the correct form is $x = 3, -3$.

Okay thank you, I seem to write it perfectly in my writing so it's all good

Okay so I would get

Sin x = 0, 90 and 180??