sin(2x)sinx=cosx?

Question

give all solutions for x in radians, between 0 and 2pi

Answer 1

we know that sin2x=2sinxcosx
therefore,
(2sinxcosx)sinx-cosx=0
cosx(2sin^2x-1)=0
cosx=0 or x=pi/2 and 3pi/2

2sin^2x-1=0

sinx=1/sqrt(2) or x=pi/4 and 3pi/4

Answer 2

We want to find all x>=0 and x <=2pi with sin(2x)sin(x)=cos(x)

We will make use of the trig identity sin(2x) = 2sin(x)cos(x)

sin2(x)sin(x)=2sin(x)cos(x)sin(x)

So we have 2sin^2(x)cos(x)=cos(x)

One solution then is cos(x)=0 which makes both sides 0.
x = Arccos(0) = 0 or 2pi

Now we assume cos(x) isn't zero to find the other solution(s)

So divide through by cos(x)

2sin^2(x)=1

sin^2(x) = 1/2

sin(x) = +sqrt(2)/2 and sinx(x) = -sqrt(2)/2 are both solutions.

x = Arcsin(sqrt(2)/2) = pi/4 or 3pi/4

x = Arcsin(-sqrt(2)/2 = 5pi/4 or 7pi/4

So a complete set of solutions for x is (0,pi/4,3pi/4,5pi/4,7pi/4,2pi)

Answer 3

sin(2x)sin x = cosx
Replace sin(2x) with its equal 2sin x*cos x to get:
2sinx cos x sin x = cos x
(sin x)^2 =1
sin x = + or - 1
Therefore x = + or - pi/2, 3pi/2, 5pi/2, 7pi/2 .......

Answer 4

sin(2x)sinx=scosx
2sinxcosxsinx=cosx
2sin^2xcosx-cosx=0
cosx(2sin^2x-1)=0
cosx=0 or x=(2n+1)pi/2
2sin^2x-1=0
2sin^2x=1
sin^2x=1/2
sinx=+/-1/rt2=sinpi/4
x=npi+(-1)^n(pi/4)
or npi+(-1)^n(-pi/4)

Answer 5

why would you want to waste your time with that question. You will never need this answer in your life...and if you do...you should already know the correct answer!