# Easy Trigonometry (5 questions)

• Apr 8th 2013, 01:37 PM
LillyJackson06
Easy Trigonometry (5 questions)
Solve each equation on the interval [0,2pi]

1) sin^2x-2cosx-2=0

2) 4cos^x=5-4sinx

3) sin2x=cosx

4) cos2x=cosx

5) cos2x+5cosx+3=0
• Apr 8th 2013, 02:07 PM
topsquark
Re: Easy Trigonometry (5 questions)
Quote:

Originally Posted by LillyJackson06
Solve each equation on the interval [0,2pi]

1) sin^2x-2cosx-2=0

2) 4cos^x=5-4sinx

3) sin2x=cosx

4) cos2x=cosx

5) cos2x+5cosx+3=0

Please show what work you have been able to do. We are not a homework service.

-Dan
• Apr 8th 2013, 02:15 PM
LillyJackson06
Re: Easy Trigonometry (5 questions)
I do not know how to start these equations, and what to substitute in, which is the first step, which i need help on.
• Apr 8th 2013, 02:27 PM
Plato
Re: Easy Trigonometry (5 questions)
Quote:

Originally Posted by LillyJackson06
Solve each equation on the interval [0,2pi]
1) sin^2x-2cosx-2=0
2) 4cos^x=5-4sinx
3) sin2x=cosx
4) cos2x=cosx
5) cos2x+5cosx+3=0

Write #1 as $1-\cos^2(x)-2\cos(x)-2=\cos^2(x)+2\cos(x)+1=0$.
Then let $u=\cos(x)$ solve $u^2+2u+1=0$.

You have left something out of #2.

#3 $\sin(2x)=2\sin(x)\cos(x)$

#5 $\cos(2x)=2\cos^2(x)-1$
• Apr 8th 2013, 03:39 PM
LillyJackson06
Re: Easy Trigonometry (5 questions)
I now understand #1.

Sorry, #2 should be 4cos^2x=5-4sinx.

I also now get #3 and #5.

Thanks so much.
• Apr 8th 2013, 03:55 PM
Plato
Re: Easy Trigonometry (5 questions)
Quote:

Originally Posted by LillyJackson06
I now understand #1.

Sorry, #2 should be 4cos^2x=5-4sinx.

I also now get #3 and #5.

#2 $4(1-\sin^2(x))=5-4\sin(x)$