# Math Help - Solve trig equatiom

1. ## Solve trig equatiom

The equation is on the interval [0,2pi]:2cos^2x+3sinx-3=0

I get up to the part -2sinx^2+3sinx-1=0

I cant get past this, im stuck as to how i should factor it.

Can domeone also tell me if i can do this on the ti-nspire cas? If so, how?

2. ## Re: Solve trig equatiom

$2\cos^2 x+3\sin x-3=0$

$2(1-\sin^2 x)+3\sin x-3=0$

$2-2\sin^2 x+3\sin x-3=0$

$-2\sin^2 x+3\sin x-1=0$

make $\sin x = u$

$-2u^2 +3u-1=0$

3. ## Re: Solve trig equatiom

Ok i had that part....i factored it as

(-sinx+1)(2sinx-1)=0

Sin x =1 which is pi/2 and 2sinx=0 which is 2pi?

Can someone please confirm?

4. ## Re: Solve trig equatiom

The second factor

$2\sin x-1 = 0$

$2\sin x = 1$

$\sin x = \frac{1}{2}$

$x = \frac{\pi}{6}$

5. ## Re: Solve trig equatiom

Whoops...thanks for pointing that out!

6. ## Re: Solve trig equatiom

Wait, somethings still wrong...

The first equation is

-sinx-1=0
And that gives me
-sinx=1 and as per the even and odd identity it is the same as sinx=-1 correct?
So what would my ans be? 3pi/2?

And for the second equation, sin x =1/2 so would my ans be pi/6+2npi and 5pi/6+2npi?

7. ## Re: Solve trig equatiom

I think i still have the factoring wrong, can someone please help?

8. ## Re: Solve trig equatiom

The equation is on the interval [0,2pi]:2cos^2x+3sinx-3=0
$-2\sin^2{x} + 3\sin{x} - 1 = 0$

$2\sin^2{x} - 3\sin{x} + 1 = 0$

$(2\sin{x}-1)(\sin{x}-1) = 0$

$\sin{x} = \frac{1}{2}$

$\sin{x} = 1$

now, pay attention to the given interval in the original problem statement and write the three solutions for x

9. ## Re: Solve trig equatiom

I see, why do i have to multiply by neg -1?

And the value for sin x=1 is pi/2+2npi? And for sin 1/2 is pi/6+2npi and 5pi/6+2npi?

I have to write the 2npi, right?

10. ## Re: Solve trig equatiom

Originally Posted by noork85
I see, why do i have to multiply by neg -1? makes it easier to factor

And the value for sin x=1 is pi/2+2npi? And for sin 1/2 is pi/6+2npi and 5pi/6+2npi?

I have to write the 2npi, right?
NO, read this again ...

The equation is on the interval [0,2pi]:2cos^2x+3sinx-3=0

11. ## Re: Solve trig equatiom

I thought that because its on the 2pi interval, it has infinitely more solutions, which is why its 5pi/6 +2npi and so on...

If not that, then i really dont know...