[MOVED] trig review: 2sin^2 (x) - 5sin (x) + 2

[velocity]

2sin^2 (x) - 5sin (x) + 2

Cany you help me with this? I have forgotten all my trig learnings. Thanks!

 

[galactus]

I assume you want to solve:

\(\displaystyle \L\\2sin^{2}({\theta})-5sin({\theta})+2=0\)

If so, let \(\displaystyle x=sin({\theta})\) and solve the resulting quadratic.

 

[velocity]

is this right?

ok. so I would have 2x^2 - 5x + 2 = 0

Ok. I subtracted the 2 and then divided by the 2
This gave me x^2 - 5x + 1 = 0

Is this right? Should I factor this now?

 

[galactus]

If you divide by 2 that middle term should be 5/2, shouldn't it?.

Just factor the way it is:

\(\displaystyle (x-2)(2x-1)\)

 

[velocity]

ok so the places where sin(x) = 1/2 are pi/3 and 5pi/3

but what about sin(x)= 2? isn't that the same as going around the unit circle 2 times, so would it be 2pi?

 

[skeeter]

-1 < sinx < 1

so, sinx = 2 has no solution.

 

[velocity]

the sin(x) = 1/2 should be pi/6 and 5pi/6 though right?

 

[skeeter]

you tell me ...

unit circles