Question

FInd all values of x
3sinx=1+cos2x

Answer 1

3*2*(sinx(x/2) )cos(x/2) - 2(cos(x/2))^2 = 0

Answer 2

2cos(x/2) *(3sin(x/2) - cos(x/2)) =0
thus cos(x/2)=0 means x/2 =(2n-1)(pi/2
or x= (2n-1)(pi)

Answer 3

can you expain how you got the first equation from the beginning equation?

Answer 4

sinx =2*sin(x/2)*cos(x/2)
and cosx= 2(cos(x/2))^2 -1

Answer 5

other values can be found from (3sin(x/2) - cos(x/2)) =0

Answer 6

@matricked im still unsure :/ im getting confused by the way its written @robtobey can you help explain? i kinda still dont unerstand

Answer 7

There is a trigonometric identity, called double angle identity. It's\[\sin(2x)=2\sin x \cos x\]\[\cos(2x)=\cos^2x-\sin^2x=2\cos^2x-1=1-2\sin^2x\]

Answer 8

Also, for tangent\[\tan(2x)=\frac{ 2\tan x }{ 1-\tan^2x }\]

Answer 9

Because there is sin(x) at the left side and cos(2x) on the right side, it would be the easiest to use the identity cos(2x)=1-2sin^2x to put everything in terms of sin(x)

Answer 10

Refer to the Mathematica attachment.

Answer 11

So the equation becomes \[3\sin x=1+1-2\sin^2x=2-2\sin^2x\]

Answer 12

\[2\sin^2x+3\sin x-2=0\rightarrow (\sin x+2)(2\sin x-1)=0\]

Answer 13

Can you take it from here?

Answer 14

@El3th1897

Answer 15

thank you for replying @science0229 and yea i make the 2 factored equations equal zero then find sinx on both to find the degrees of sinx right?

Answer 16

yeah.

Answer 17

the first one it becomes sin(x)=-2 there is no solution for that right?

Answer 18

@science0229

Answer 19

yeah

Answer 20

cool

Answer 21

you just have to find the solution for sin(x)=1/2.

Answer 22

I'm sure that you can do it, right?

Answer 23

\[\Pi/6 and 5\Pi/6 \right?\]

Answer 24

thats in radians pi/6 and 5pi/6

Answer 25

over the interval of [0,2pi].

Answer 26

have to eat. sorry.
different time zone here( seoul time zone)

fyi. the general solution is (pi/6+2npi) and (5pi/6+2npi) where n=integer