Trig question

Question

Trig question

joannablackwelder · Answer

attachment

joannablackwelder · Answer

@TheSmartOne Can you help with this one?

joannablackwelder · Answer

I get no solution. But apparantly, that isn't right.

thesmartone · Answer

We need to use this identity.
cos(2x) = 1 – 2sin^2(x)

thesmartone · Answer

So:

$\sf 6-sin(\theta) = 1 - 2sin^2\theta$

$\sf 2sin^2 \theta-sin \theta + 6 - 1 = 0$

$\sf 2sin^2 \theta -sin\theta + 5 = 0$

thesmartone · Answer

Let's say $\sf sin\theta = x$

Then we need to factor

$\sf 2x^2 - x + 5 = 0$

joannablackwelder · Answer

Hm, I see. So, why do I not get the right answer when I graph both of them on my calc and they don't cross?

joannablackwelder · Answer

But even that method gives imaginary solutions.

thesmartone · Answer

To be honsest, I don't have a graphing calculator, so I had to learn how to do it manually. :P

joannablackwelder · Answer

Oh, ok, but don't you get imaginary solutions when you plug that into the quadratic?

thesmartone · Answer

I do, mhmmm. I guess they want the imaginary solutions to the question?

joannablackwelder · Answer

Hm, seems very strange.

joannablackwelder · Answer

Any thoughts, @Nnesha ?

nnesha · Answer

i'll go with no solution 
but ..hm  i'm not sure sorry

nnesha · Answer

@jim_thompson5910

joannablackwelder · Answer

Ok, thanks!