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?
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?
$\displaystyle 2\cos^2 x+3\sin x-3=0$
$\displaystyle 2(1-\sin^2 x)+3\sin x-3=0$
$\displaystyle 2-2\sin^2 x+3\sin x-3=0$
$\displaystyle -2\sin^2 x+3\sin x-1=0$
make $\displaystyle \sin x = u $
$\displaystyle -2u^2 +3u-1=0$
factor the quadratic
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?
$\displaystyle -2\sin^2{x} + 3\sin{x} - 1 = 0$The equation is on the interval [0,2pi]:2cos^2x+3sinx-3=0
$\displaystyle 2\sin^2{x} - 3\sin{x} + 1 = 0$
$\displaystyle (2\sin{x}-1)(\sin{x}-1) = 0$
$\displaystyle \sin{x} = \frac{1}{2}$
$\displaystyle \sin{x} = 1$
now, pay attention to the given interval in the original problem statement and write the three solutions for x