Question

quick question!

Answer 1

question?

Answer 2

solutions in (0,2pi)

Answer 3

this is a long process u ready?

Answer 4

yeah i am.

Answer 5

1-2sin^2(x)+root 2sinx=1
-root 2sinx(root2sinx+1)=1
root2sinx+1=1
root2sinx=0...
idk well...lol

Answer 6

lol ok ill try your answer thanks!

Answer 7

cos(2x)=1-2sin^2(x)
do u still need help or gunna go with his answer?

Answer 8

i still need your help!

Answer 9

wow...wow....stop...that's not the ans...u proceed urself...

Answer 10

so can u simplify the thing i gave u?

Answer 11

is there a multiple choice?

Answer 12

nope

Answer 13

so simplify

Answer 14

ok how?

Answer 15

ok so now u have..

Answer 16

\[2\sin^2(x)+\sqrt{2}\sin(x)=0\]

Answer 17

hello?

Answer 18

huh

Answer 19

\[\cos2x =1-2\sin^2x\] given\[\cos2x+\sqrt{2}sinx=1 \]\[1-2\sin^2x+\sqrt{2}sinx=1\]\[2\sin^2x+\sqrt{2}sinx=0\]\[\sqrt{2}sinx(\sqrt{2}sinx+1)=0\]\[sinx=0\]\[or\]\[\sqrt(2)sinx=-1\]\[sinx=(-1/\sqrt(2))\]which is \[x= 5\pi/4\] so x values lies between \[(0,2\pi)\]

Answer 20

exactly what i was trying to say

Answer 21

wow..u hv done it. good good.

Answer 22

its ok @jessiegonzales

Answer 23

great job though @naveenrockzz

Answer 24

its not that much difficult @jessiegonzales @Probhat

Answer 25

LOL i know i was just trying to explain but ppl these days don't want to work it out they just want the answer... @naveenrockzz

Answer 26

ha that's true @jessiegonzales