Question

Find solutions for 2cos^2(theta)-sin(theta) =1 from [0,2pi)

Answer 1

method is to rewrite \[cos^2(t) = (1-sin^2(t)\]

Answer 2

\[2(1-sin^2(t))-sin(t)=1\]
\[2-2sin^2(t)-sin(t)=1\]
\[2sin^2(t)+sin(t)-1=0\]

Answer 3

now you have a quadratic equation in \[sin(t)\]

Answer 4

like solving
\[2x^2+x-1=0\]

Answer 5

where \[x=sin(t)\]
\[2x^2+x-1=0\]
\[(2x-1)(x+1)=0\]
\[x=-1\] or \[x=\frac{1}{2}\]

Answer 6

so now solve
\[sin(t)=-1\] means
\[t=\frac{3\pi}{2}\]

Answer 7

\[sin(t)=\frac{1}{2}\]
\[t=\frac{\pi}{6}\] or \[t=\frac{5\pi}{6}\]

Answer 8

done!

Answer 9

ty

Answer 10

welcome!