Question

hey,I need to solve the equation

\[2\frac{ sinx }{ 2 }*\frac{ cosx }{ 2}*cosx=\frac{ 1 }{ 2 }\]
\[cosxsinx*\frac{ cosx }{ 2 }=\frac{ 1 }{ 2 }|*2\]
\[2sinxcosx*cosx=1\]
\[\sin(2x)cosx=1\]

Did I do anything wrongly? If no,what to do next?

Answer 1

@hartnn

Answer 2

What was the initial problem?

Answer 3

I need to solve the equation 2*sinx/2*cosx/2*cosx=1

Answer 4

sorry

Answer 5

2*sinx/2*cosx/2*cosx=1/2

Answer 6

\[2\sin(\frac{x}{2})\cos(\frac{x}{2})\cos(x)=\frac{1}{2} \\ \text{ or } \\ 2\frac{\sin(x)}{2}\frac{\cos(x)}{2}\cos(x)=\frac{1}{2}\]

Answer 7

\[2\sin(\frac{x}{2})\cos(\frac{x}{2})\cos(x)=\frac{1}{2} \\ \text{ Recall } \sin(x)=2 \sin(\frac{x}{2}) \cos(\frac{x}{2}) \\ \sin(x)\cos(x)=\frac{1}{2} \\ 2\sin(x)\cos(x)=1 \\ \sin(2x)=1 \]

Answer 8

That should help...
The other thing I could probably give you a method that will lead to some extraneous answers.

Answer 9

sin(2x)=1
let u=2x
find when sin(u)=1 (use unit circle)

Answer 10

oh,okay,thanks. I probably understood the problem incorrectly:)