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Help with trigonometric equations?

Hi everyone,

I'm sorry if this thread is in the wrong place, it seemed like the best option. I've been trying so hard to answer this question and I can't seem to get past the first step.

Any help would be greatly appreciated, an explanation would be particularly helpful.

1) Solve sin2x/4sin2x-1=0
For 0<=x<180

2) Solve 5sin^2x-2cosx-5=0
For 0<=x<2pi
(edited 9 years ago)
Original post by Mozzmo
Hi everyone,

I'm sorry if this thread is in the wrong place, it seemed like the best option. I've been trying so hard to answer this question and I can't seem to get past the first step.

Any help would be greatly appreciated, an explanation would be particularly helpful.

1) Solve sin2x/4sin2x-1=0
For 0<=x<180


Could you clarify what the fraction is
Reply 2
Avatar for Mozzmo
Mozzmo
OP
2
Original post by TenOfThem
Could you clarify what the fraction is


It is Sin2x over 4sin2x-1 =1

Sorry about that.
Original post by Mozzmo
It is Sin2x over 4sin2x-1 =1

Sorry about that.


Multiply both sides by 4Sin2x - 1

Should be easy then
Original post by Mozzmo


2) Solve 5sin2x-2cosx-5=0
For 0<=x<2pi


This one is unusual are you sure it isn't 5Sin2x - 2Cosx - 5 = 0
Reply 5
Avatar for Mozzmo
Mozzmo
OP
2
Original post by TenOfThem
Multiply both sides by 4Sin2x - 1

Should be easy then



Thanks but I managed to get that far and then couldn't seem to understand what to do next. I tried to rearrange but I got a math error.

For the second question, yes, you're correct, I typed it out wrong. I'll edit the post.
Original post by Mozzmo
Thanks but I managed to get that far and then couldn't seem to understand what to do next. I tried to rearrange but I got a math error.


From Sin2x = 4Sin2x - 1

The next line is

3Sin2x = 1


That should be fairly straightforward
(edited 9 years ago)
Original post by Mozzmo


For the second question, yes, you're correct, I typed it out wrong. I'll edit the post.


Thought so

Use Sin^2x = 1 - Cos^2x

Then you have a quadratic
Reply 8
Avatar for Mozzmo
Mozzmo
OP
2
Original post by TenOfThem
Thought so

Use Sin^2x = 1 - Cos^2x

Then you have a quadratic


Thank you, I understand both now. I appreciate the help.
Reply 9
Avatar for Jgoodwin
Jgoodwin
1
Original post by TenOfThem
Thought so

Use Sin^2x = 1 - Cos^2x

Then you have a quadratic


It then becomes 5cos^2x + 2cosx = 0 and that can be written as 5y^2 + 2y for ease......
But how do i then utilise that for radians??
let y = sin2x then it should be easier(dont forget to sub sin2x back in!)
Reply 11
Avatar for davros
davros
16
Original post by bellesox
let y = sin2x then it should be easier(dont forget to sub sin2x back in!)

Er, you're replying to a thread that is 5 years old!! :smile:

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