Question

4 sin θ cos θ + 2 cos θ − 2 sin θ − 1 = 0

Answer 1

Find all solutions of the equation.

Answer 2

how do i factor this?

Answer 3

I dont see any sin 2θ

Answer 4

@nathan917

Answer 5

Yes, you do.

Answer 6

from the first two terms, factor out 2*cos(t)
and from other two, (-1)
\[2\cos\theta(2\sin\theta+1)-(2\sin\theta+1)=0\]

Answer 7

\[2(2\sin\theta \cos \theta) + 2\cos \theta - 2\sin \theta - 1 = 0\]

Answer 8

now factor the \[2\sin\theta+1\] you get
\[(2\sin\theta+1)(2\cos\theta-1)=0\]

Answer 9

Eh, yes. But I was just pointing out that there's a \(\sin(2 \theta)\) in there.

Answer 10

I'm lost

so

You can express it in a simple fashion as factorized

(2 sin(θ) + 1) * (2 cos(θ) -1) = 0

and the solutions are given from sin (θ) = -1/2 and cos(θ) = 1/2, which in the interval (0, 2π) are

θ = (5/3)pi, (1/3)pi, (11/6)pi, (7/6)pi

It is periodic in θ with period (2pi), so the general solutions are given from the above adding 2pi*N, N being an integer?

Answer 11

yep. that seems the easiest way to get the solution

Answer 12

As i said b4 nut took it off though it was rong :/