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

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- Apr 8th 2013, 12:37 PMLillyJackson06Easy 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, 01:07 PMtopsquarkRe: Easy Trigonometry (5 questions)
- Apr 8th 2013, 01:15 PMLillyJackson06Re: 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, 01:27 PMPlatoRe: Easy Trigonometry (5 questions)

Write #1 as $\displaystyle 1-\cos^2(x)-2\cos(x)-2=\cos^2(x)+2\cos(x)+1=0$.

Then let $\displaystyle u=\cos(x)$ solve $\displaystyle u^2+2u+1=0$.

You have left something out of #2.

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

#5 $\displaystyle \cos(2x)=2\cos^2(x)-1$ - Apr 8th 2013, 02:39 PMLillyJackson06Re: 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, 02:55 PMPlatoRe: Easy Trigonometry (5 questions)