Huge help in Trigonometry needed!

[Hiep HO]

Hi Guys!

I came across this question that baffled ma mind...
Would appreciate it if you kind chaps could help me out?


=========Typed exactly from my sheet=========

3. Find values for x in the range 0 ≤ x ≤ π.

(i) 2sinx cosx = sinx
(i) 2sin²x - cosx -1 = 0

--------------------------------------
==KEY==
π - Pi
--------------------------------------

Thanks to everyone who particated I stay alive for 1 more day
Well appreciated!

 

[malyce]

More algebra/trig! YAY!

Easy peasy, baby. Do you remember your algebra rules?

So we have:
(i) 2sinx cosx = sinx

There's a sinx on both sides. What's that tell you? It tells me they're going to cancel each other out if we divided both sides by sinx.

2cosx = 1. After we divide both sides by sin x. What's next you ask! Why, I'll tell you! Lets get that cosx alone.

cosx = 1/2. (dividing both sides by 2).

Remember the last homework problem where you had to solve for theta? x is your theta. x = cos^-1(1/2). Now that you have an angle for x, divide 360 by it and you get your fraction for how it relates to 2*pi.

Can you do the second question by yourself? Here's a hint for you: Sin^2x - 1 = a trig identity. That = -cos^2x. So your problem then becomes:

-2cos²x - cosx = 0

Rewrite it to -2cos²x = cosx.
Divide by cosx. -2cosx = 1.
cosx = -1/2. Look at that, same as the question above.

 

[Koot]

Malyce, you need to check your algebra.

2sin²x - 1 =/= -2cos²x

This would only be true if the -1 was a -2.

 

[malyce]

Malyce, you need to check your algebra.

2sin²x - 1 =/= -2cos²x

This would only be true if the -1 was a -2.

you're right, that's my fault. I'll resolve this later today. That's what I get for doing math on 56 hours of no sleep.

It should turn into 2sinx = cosx/sinx + arcsinx

2sinx = arctanx+arcsinx

 

[Koot]

I don't know what you're doing but if you simplify it, then what I got was.

2cos²x + cosx - 1 = 0

Then you can treat it as a quadratic equation.

 

[malyce]

I don't know what you're doing but if you simplify it, then what I got was.

2cos²x + cosx - 1 = 0

Then you can treat it as a quadratic equation.

That doesn't work as a quadratic. If it was cos²x + cosx - 1 = 0 it would be, or
cos²x + 2cosx + 1 = 0.

This is just an ugly problem all around.

 

[malyce]

I don't know what you're doing but if you simplify it, then what I got was.

2cos²x + cosx - 1 = 0

Then you can treat it as a quadratic equation.

That doesn't work as a quadratic. If it was cos²x + cosx - 1 = 0 it would be, or
cos²x + 2cosx + 1 = 0.

This is just an ugly problem all around.

Wait, my brain derped. You can apply the quadradic formula to this, but you still won't get a single sign or cosine value for x.

 

[malyce]

I don't know what you're doing but if you simplify it, then what I got was.

2cos²x + cosx - 1 = 0

Then you can treat it as a quadratic equation.

That doesn't work as a quadratic. If it was cos²x + cosx - 1 = 0 it would be, or
cos²x + 2cosx + 1 = 0.

This is just an ugly problem all around.

Wait, my brain derped. You can apply the quadradic formula to this, but you still won't get a single sin or cosine value for x.

 

[Koot]

Yes you will... I'm not even sure what you're doing anymore.

If you sub in cosx for say z

Then it'll be 2z² + z - 1 = 0

This can easily be factorable into (2z - 1)(z + 1) = 0

Sub back in cosx and now you can easily find out the values of x.