# some more trig identities

• Sep 4th 2010, 02:57 PM
jayshizwiz
some more trig identities
hi. I think i'm getting the hang of these problems but I still need some assistance:

1) $\displaystyle cos^2x+sinx=1$
$\displaystyle cos^2x+sinx=cos^2x+sin^2x \rightarrow sinx - sin^2x=0 \rightarrow sinx(1-sinx)=0$
$\displaystyle sinx_1 = 0 \rightarrow x_1=180n$
$\displaystyle 1-sinx_2=0 \rightarrow sinx_2=1 \rightarrow x_2=90+360n$

2) $\displaystyle 4sin^3x-sinx=0$
$\displaystyle sinx(4sin^2x-1)=0$
$\displaystyle sinx_1 = 180n$
$\displaystyle 4sin^2x_2-1=0 \rightarrow sin^2x=1/4 \rightarrow sinx=1/2 \rightarrow x_2 = +-30+180n$

3) $\displaystyle 2cos^2x-5sinx-4=0$
$\displaystyle 2-2sin^2x -5sinx-4=0$ ???

are 1 and 2 correct? number 2 I wasn't sure about (sinx)^2=1/4 - - - sinx=1/2. can i just do a square root like that? and how do I solve 3?

Thanks
• Sep 4th 2010, 04:50 PM
Quote:

Originally Posted by jayshizwiz
hi. I think i'm getting the hang of these problems but I still need some assistance:

1) $\displaystyle cos^2x+sinx=1$
$\displaystyle cos^2x+sinx=cos^2x+sin^2x \rightarrow sinx - sin^2x=0 \rightarrow sinx(1-sinx)=0$
$\displaystyle sinx_1 = 0 \rightarrow x_1=180n$
$\displaystyle 1-sinx_2=0 \rightarrow sinx_2=1 \rightarrow x_2=90+360n$

2) $\displaystyle 4sin^3x-sinx=0$
$\displaystyle sinx(4sin^2x-1)=0$
$\displaystyle sinx_1 = 180n$
$\displaystyle 4sin^2x_2-1=0 \rightarrow sin^2x=1/4 \rightarrow sinx=1/2 \rightarrow x_2 = +-30+180n$

3) $\displaystyle 2cos^2x-5sinx-4=0$
$\displaystyle 2-2sin^2x -5sinx-4=0$ ???

are 1 and 2 correct? number 2 I wasn't sure about (sinx)^2=1/4 - - - sinx=1/2. can i just do a square root like that? and how do I solve 3?

Thanks

(1) is good...

For (2), you neglected to write the negative square root of 1/4, which is -1/2, but you found the angles nonetheless.

(3) can written as a quadratic equation, using

$\displaystyle -2Sin^2x-5Sinx-2=0$

$\displaystyle 2Sin^2x+5Sinx+2=0$

$\displaystyle (2Sinx+1)(Sinx+2)=0$

and work from there
• Sep 4th 2010, 11:55 PM
jayshizwiz
thanks. i can't believe i didn't realize that. here's what I did...

http://www.mathhelpforum.com/math-he...7495a6ea54.png

$\displaystyle sinx(2sinx+5)=-2$ and then i was stuck...lol