Question

Use an identity to solve each equation on interval [0,2pi)

2cos2x-sinx-1=0

Answer 1

Is this your question?\[2\cos(2x)-sinx-1=0?\]

Answer 2

yes

Answer 3

Ok we have to solve this by factoring, and we need to change the cosine into sine to match it, so we can use the identity \(\cos(2x)=1-2\sin^2(x)\)
\[2(1-2\sin^2(x))-\sin(x)-1=0 \\ \\ 2-4\sin^2(x)-\sin(x)-1=0 \\ \\ -4\sin^2(x)-\sin(x)+1=0\]
We can factor it now, can you continue?

Answer 4

No this is the part i suck at the mostt :(

Answer 5

Oh when you can't factor it, use quadratic formula.

Answer 6

\[-4\sin^2(x)-\sin(x)+1=0 \\ \\ Ax^2+Bx+C=0 \\ \\ \text{When} \\ \\ A=-4, B=-1, C=1\]
\[\sin(x)=\frac{-B \pm \sqrt{B^2-4AC}}{2A}\]

Answer 7

so it will be:
\[\frac{ -1\pm \sqrt{-15} }{ -8 }\]

Answer 8

Careful

Answer 9

\[\sin(x)=\frac{-(-1) \pm \sqrt{(-1)^2-4(-4)(1)}}{2(-4)}\]

Answer 10

Positive 1

Answer 11

\[\sin(x)=\frac{1 \pm \sqrt{17}}{-8}\]

Answer 12

Then, we have 2 sines out of this
\[\sin(x)=-\frac{1}{8}+\frac{\sqrt{17}}{8}~~~~~~or~~~~~~\sin(x)=-\frac{1}{8}-\frac{\sqrt{17}}{8}\]

Answer 13

Inverse that will give you x
\[x=\sin^{-1}(-\frac{1}{8}+\frac{\sqrt{17}}{8})~~~~~or~~~~~x=\sin^{-1}(-\frac{1}{8}+\frac{\sqrt{17}}{8})\]

Answer 14

So after i find x how will i solve for the intervals between [0,2pi)

Answer 15

Use a calculator to find that, you should get
x=22.97 and 39.8

Answer 16

For positive angle, look at the positive side
x=180-22.97=157
For negative angle is the negative side
x=180+39.8=219.8
x=320.2

So we got
x=22.97,157,219.8,320.2

39.8 doesn't include because its negative

Answer 17

and also, you have to convert the angles to radians

Answer 18

In my homework they have it in radians. I should just convert?

Answer 19

\[\frac{\theta}{180} \times \pi=randians\]
\[\frac{radians}{\pi} \times 180=degrees\]