Question

what is the trig solution to 2cosxsinx-cosx=0

Answer 1

you have 2 ways
1. \[\cos x(2\sinx-1)=0\]
2. \[\sin {2x}=\sin(\frac{\pi}{2}-x)\]

Answer 2

@Disco94 , you got it?

Answer 3

what???

Answer 4

Sorry. :)

Answer 5

Factor out cos x
\[\cos x(2\sin x-1)=0\]
\[\cos x=0 \]
or
\[\sin x=\frac{1}{2}\]

Answer 6

\[ 2 \cos x \sin x- \cos x=0 \implies \cos x (2 \sin x -1) = 0 \]

\[\cos x = 0 \implies x = (2n+1) \frac \pi 2, n \in \mathbb{W} \]

or \[ \sin x = \frac 12 \implies x = n\pi +(-1)^n \frac \pi 6, n \in \mathbb{W} \]

Answer 7

the answer is the cosx= pie/2 and 3pie/2 the sinx=1/2 and pie/6 and 5pie/6 but i need to know how to get the answer

Answer 8

So the simple answer is
\[x=\frac{pi}{2}, \frac{3pi}{2}, \frac{pi}{6}, \frac{5pi}{6}+2n pi\]

Answer 9

Put \( n = 0,\pm 1 \) in the equation in my derivation you will get your desired answer. But I have to say the question is not clear.