Numeric Math Problem

Question

Solve Sin2x-1= cos2x for values of x in the interval 0  ≤  x  ≤  2pi

Answer 1

sin 2x - 1 = cos 2x

2 sin x cos x - 1 = 2 (cos x)^2 - 1 <---- trig identities for sin 2x and cos 2x

2 sin cos - 1 = 2 cos^2 - 1 <----- angle argument dropped since trig functions contain the same angle

so as to make notation easier to type

2 sin cos = 2 cos^2 <--- adds 1 to both sides

sin cos = cos^2 <--- divides both sides by 2

0 = cos^2 - sin cos <--- subtracts sin * cos from both sides

0 = cos ( cos - sin) <--- factors out cos

cos = 0 or cos - sin = 0

working with the right equation....

cos = sin <---- adds sin to both sides

cos = sqrt(1 - cos^2) <--- trig identity: sin^2+cos^2=1

cos^2 = 1 - cos^2 <--- squares both sides

2cos^2 = 1

cos^2 = 1/2

cos = +or-sqrt(2)/2

So cos=0 or cos = +or- sqrt(2)/2

the angle then must be pi/2, 3*pi/2, pi/4, 3*pi/4, 5*pi/4, 7*pi/4

in degrees, they are 90, 270, 45, 135, 215, 315