1-sin2x = cosx- sinx 0<x<2pi solve for x?

Question

please answer today....due in tomorow.
thankyou

There are supposedly 5 answers?
But I'm unsure about it.

But there is definitly more than one

Answer 1

1-sin2X = cos^2(x) + sin^2(x) - 2sin(x)cos(x)
= [cos(x)-sin(x)]^2
Problem says [cos(x)-sin(x)]^2=cos(x)-sin(x)
So either cos(x)-sin(x)=0 ie x= pi/4 or 5pi/4
Or cos(x) - sin(x) = 1 which means cos(x) = 1 + sin(x)
In this case sin(x) is clearly negative. Solve this yourself.

Answer 2

1- sin 2x = sin^x + cos^x - 2sin x cos x = (cosx - sin x)^2
(using sin ^ x+ cos ^x =1, sin 2x = 2 sinx cos x and then
(a-b)^2 = a^2 -2ab +b^2

now (cosx - sin x)^2= cos x - sin x
so cos x - sinx = 0 or cos x - sin x = 1

(using y^2 = y => y^2-y = 0 =>y(y-1) = 0
cos x -sin x = 0=>tan x = 1 so x = pi/4 or 5pi/4(1st and 3rd quadrant)

cos x - sin x = 1

LHS is of the form a cos x + b sin x( a = 1 and b = -1)

if a = r cos y and b = r sin y
r = 2^(1/2), y = -pi/4
LHS = r cos y cos x + y sin y sin x
= r cos(x+y) = 1
cos(x+y) = 1/r = 1/2^1(1/2) = cos pi/4
x +y = pi/4 or -pi/4
x = pi/4 + y or -pi/4 +y
= pi/4-pi/4 or -pi/4 -pi/4
= 0 or -pi/2 =0 or 3pi/2
(add 2pi to make positive proper quadrant)
so we have solution 0, pi/4,5pi/4 3pi/2

Answer 3

1-2sinxcosx=cosx-sinx
(cos-sinx)^2=cosx-sinx
cosx+sinx=1
x=0 so cos0+sin0=1
x=135* cos0=-1/rt2 sin 135*=sin45*=1/rt2
solution is x=135*

Answer 4

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RE:
1-sin2x = cosx- sinx 0<x<2pi solve for x?
please answer today....due in tomorow.
thankyou

Answer 5

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use cos( A-B) formula it goes : cos(A - B) = cos AcosB + sinAsinB so cosxcos2x + sinxsin2x = 1/sqrt(2) ie cos(2x-x) = 1/sqrt(2) we know that cos(pi/4)=1/sqrt(2) so, x = pi/4 and also 2pi-pi/4 ie x= pi/4, 7pi/4

Answer 6

1-sin 2x

Answer 7

Well, I don't know; friginometry eludes me but at least your avatar doesn't look as stupid and fruity as the two other guys who answered you. Especially the one that's smiling.