Question

please help!!! will give medal! its a trig proof

Answer 1

\[\sin(2x) + \cos(3x) = 0\]

Answer 2

Well this is not a proof

Answer 3

This is an equation

Answer 4

well i mean prove it

Answer 5

It can't be proven, it's not always true

Answer 6

You mean solve it?

Answer 7

well, if you were to attempt it, what would you do?

Answer 8

So it's proof it or solve it -_-

Answer 9

i guess it's solving

Answer 10

Subtract sin(2x) from both sides

Answer 11

okay \[\cos(3x) = -\sin(2x)\]

Answer 12

Put the negative inside the sine

Answer 13

so\[\cos(3x) = (-\sin(2x))\]

Answer 14

I mean \(\cos(3x)=\sin(-2x)\)

Answer 15

You now have two options

Answer 16

Either you change cosine to sine, or you change sine to cosine.

Answer 17

\[\begin{array}{rcl}
\sin(2x)+\cos(3x)&=&0\\
\cos(3x)&=&-\sin(2x)\\
\cos(3x)&=&\sin(-2x)\\
\cos(3x)&=&\cos(90^\circ+2x)\\
3x&=&90^\circ+2x\\
x&=&90^\circ
\end{array}\]

Answer 18

could that be written in terms of \[\pi \]

Answer 19

\[\begin{array}{rcl}
\sin(2x)+\cos(3x)&=&0\\
\cos(3x)&=&-\sin(2x)\\
\cos(3x)&=&\sin(-2x)\\
\cos(3x)&=&\cos(\frac\pi2+2x)\\
3x&=&\frac\pi2+2x\\
x&=&\frac\pi2
\end{array}\]

Answer 20

What's the range of x?

Answer 21

the question asks for the answers of x to be between (0, 2pi

Answer 22

\[\begin{array}{rcl}
\sin(2x)+\cos(3x)&=&0\\
\cos(3x)&=&-\sin(2x)\\
\cos(3x)&=&\sin(-2x)\\
\cos(3x)&=&\cos(\frac\pi2+2x)\\
3x=\frac\pi2+2x&or&3x=2\pi-\frac\pi2-2x\\
x=\frac\pi2&or&5x=\frac{3\pi}2\\
x=\frac\pi2&or&x=\frac{3\pi}{10}
\end{array}\]

Answer 23

\[\begin{array}{rcl}
\sin(2x)+\cos(3x)&=&0\\
\cos(3x)&=&-\sin(2x)\\
\cos(3x)&=&\sin(-2x)\\
\cos(3x)&=&\cos(\frac\pi2+2x)\\
3x=\frac\pi2+2x&or&3x=2\pi-\frac\pi2-2x&or&\cdots\\
x=\frac\pi2&or&5x=\frac{3\pi}2&or&5x=\frac{7\pi}2\\
x=\frac\pi2&or&x=\frac{3\pi}{10}&or&x=\frac{7\pi}{10}&or&\cdots
\end{array}\]

Answer 24

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=sin(2x)%2Bcos(3x)%3D0