Question

4 sin θ cos θ + 2 sin θ − 2 cos θ − 1 = 0
Find all solutions of the equation

Answer 1

So far I have:
\[2\sin \theta=1\]
\[\sin \theta=\frac{ 1 }{ 2 }\]
\[2\cos \theta=-1\]
\[\cos \theta=-\frac{ 1 }{ 2 }\]

Not sure what do with it next

Answer 2

\[4\sin \theta \cos \theta +2\sin \theta -2\cos \theta -1 = 0\]
\[2(2\sin \theta \cos \theta) +2\sin \theta -2\cos \theta -1 = 0\]
\[2(\sin 2\theta) +2\sin \theta -2\cos \theta -1 = 0\]

Answer 3

you can solve this by grouping

Answer 4

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

2sin t ( 2 cos t + 1) - ( 2 cos t + 1 ) = 0

Answer 5

you cant type in this program

Answer 6

Haha yeah I didn't get any of that :P

Answer 7

4 sin t cos t + 2 sin t - 2 cos t - 1 = 0

2sin t ( 2 cos t + 1) - ( 2 cos t + 1 ) = 0

Answer 8

does that make sense so far?

Answer 9

Yes it does

Answer 10

4 sin t cos t + 2 sin t - 2 cos t - 1 = 0

2sin t ( 2 cos t + 1) - ( 2 cos t + 1 ) = 0

2sin t ( 2 cos t + 1) - 1*( 2 cos t + 1 ) = 0

(2 sint - 1 ) ( 2 cos t + 1) = 0

Using zero product property

2 sin t -1 = 0 , 2cos t + 1 = 0

Answer 11

2sinθ=1

sinθ=1/2

2cosθ=−1

cosθ=−1/2

But then how do I get it so it's like 2(Pi)k+something(Pi)/something

If that makes sense.

Answer 12

Here I'll attach a screen shot of the example problem. Left side of picture is example problem. Green boxes have the correct answers in them.