Question

Find all solutions on the interval [0,2pi]

2 cosx sinx - cos x =0

Please show+explain work?

Answer 1

did you get this yet? if not i will solve.

Answer 2

go ahead :)

Answer 3

first i would write
\[2cos(x)sin(x)=cos(x)\]
then square to get
\[4cos^2(x)sin^2(x)=cos^2(x)\]
\[4cos^2(x)[1-cos^2(x)]=cos^2(x)\]
\[4cos^2(x)-4cos^4(x)-cos^2(x)=3cos^2(x)-4cos^4(x)=0\]

Answer 4

how am i doing so far?

Answer 5

If i do this I will move the cos x to the left and factor out the cos x

Answer 6

\[4cos^4(x) - 3cos^2(x)=0\]
\[cos^2(x)(4cos^2(x)-3)=0\]
\[cos^2(x)=0\] or \[cos^2(x)=\frac{3}{4}\]

Answer 7

\[cos(x) = 0\]
\[x=\frac{\pi}{2}\]or \[x=-\frac{\pi}{2}\]

Answer 8

proof 2 now that i am on a roll.

\[f'(x)=3ax^2-4ax-8a\] the zeros are
\[\frac{4a\pm\sqrt{16a^2+96a}}{6a}=\frac{4a\pm4\sqrt{a^2+6a}}{6a}=\]
\[\frac{2\pm 2\sqrt{a(a+6)}}{3a}\]
and in order for this to be -1
\[a(a+6)\] must be a perfect square, a miracle if ever there

Answer 9

ignore that last gibberish it is wrong and not from here anyway.

Answer 10

Alternative solution
\(\cos x(2\sin x -1)=0\)
\(\cos x=0\Rightarrow x=\pi/2,3\pi/2\)
\(\sin x=1/2\Rightarrow x=\pi/6,5\pi/6\).

Answer 11

btw we knew
\[\pm\frac{\pi}{2}\]worked to begin with by inspection.

Answer 12

right, much swifter and i retire now. yikes.

Answer 13

Thanks!