Question

stepwise solution: Find all the values of x in the interval [0,2pi) that satisfy the equation 2sinxcosx=cosx

Answer 1

I don't know what is meant by stepwise function here as trigonometric functions are continuou over -infinity to +infinity domain.

Answer 2

meaning how to solve this equation in steps

Answer 3

i know what is stepwise function. but here this word doesn't value.

well i try to solve this according to my knowledge. you watch and ask me if you can't get any step. i am solving.

Answer 4

\[\large \text{original funtion is} \]\[\large 2sinxcosx=cosx\]\[\large 2sinx\cancel{cosx}=\cancel{cosx}\]\[\large 2sinx=1\]\[\large sinx=\frac{1}{2}\]\[\large x=\sin ^{-1}(\frac{1}{2})\]until this step have you got this?

Answer 5

yes.

Answer 6

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

so \(\cos(x)=0\) or \(2\sin(x)-1=0\)

Answer 7

that would be pi/6

Answer 8

your answer will be.\[\large x=\sin ^{-1}(\frac{1}{2}+2n \pi)\]where n is a positive integer and its interval is\[\large [0,2 \pi)\]

Answer 9

@Zarkon tell me where did i do wrong?

Answer 10

cos(x) could be zero...in that case you can't divide by cos(x)

Answer 11

so then what are are the values x can be knowing that it's within the interval (0,2pi)

Answer 12

\[\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}\]

Answer 13

how did u get that?

Answer 14

i got\[\Pi/6 \] only

Answer 15

solve the two equations I gave you..

\(\cos(x)=0\) or \(2\sin(x)−1=0\)

Answer 16

oh ok! got it! Thank you