Question

solve the following equation for x where
0 13 years ago

Answer 1

hint: \(sin2x = 2sinxcosx\)

Answer 2

yes and

Answer 3

well where are you stuck
or show me what you did

Answer 4

the equation is not solved

Answer 5

\[ 2sinxcosx = cosx => 2sinxcosx - cosx = 0 \]
why dont you try and solve it

Answer 6

take cos(x) you will have
\[\Large \cos(x)[2\sin(x)-1]=0\]
now you have two equations
\[\Large \implies \cos(x)=0\]
\[\Large \implies 2\sin(x)-1=0\]

solve these two equations.

Answer 7

if i knew how to solve these equations i wouldn;t be asking for help

Answer 8

to this stage i know, but how to solve from here?

Answer 9

\[\cos(x) = 0 => cosx = \cos\left(\frac{\pi}{2}\right) => x = \frac{\pi}{2} +n*2\pi , n \epsilon\mathbb{Z} \]
\[sinx = \sin\left(\frac{\pi}{6}\right) => x = \left(-1\right)^{n}*\frac{\pi}{6} +n*\pi , n \epsilon\mathbb{Z} \]

Answer 10

i've been to wolfram alpha to
mimi

Answer 11

lol i didnt use wolfram..

Answer 12

the answer is correct though but i have to show it using the unit circle, or the graph

Answer 13

i think using the sin wave