Question

Solve the trigonometric equation on the interval [0,2pi)

2sin^2(theta)-cos(theta)-1=0

I've got no clue how to start this, please help.

Answer 1

rewrite \(\sin^2(\theta)\) as \(1-\cos^2(\theta)\) and then you have a quadratic equation with only cosines

Answer 2

\[2(1-\cos^2(\theta))-\cos(\theta)-1=0\] etc
you good from there?

Answer 3

I should get cos^2θ-cosθ-(1/2)=0 right?

Answer 4

Please help

Answer 5

oh sorry lost you

Answer 6

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

Answer 7

easier to leave the 2 there, it factors nicely

Answer 8

you get \[\cos(x)=-1\] or
\[\cos(x)=\frac{1}{2}\] and you can solve those

Answer 9

You are amazing, thank you sincerely.