cos2x=2tan2x?

Question

what's all the answer within [0,2pi]?

thanks.

Answer 1

cos2x =2sin2x/cos 2x so
cos^2 2x= 2sin2x and
1-sin^2 2x = 2 sin 2x.
If you call sin 2x = z
1-z^2=2z
so z^2+2z-1=0 and z = ((-2+-sqrt8))/2 =-1+-sqrt2
-1-sqrt2<-1 so can´t be
z=-1+sqrt2=0.4142=sin2x
2x=0.4271+2kpi and 2x=pi-0.4271+2kpi rad

x=0.2135+k pi So in the interval we get x=0,2135 and 0.2135+pi
x=pi/2-0.2135 +kpi so in the interval we get
x=pi/2-0.2135 and x= 3pi/2-0.2135 all in radians

Answer 2

cos2x-2tan2x = 0
(1/cos2x)(cos^2 2x - 2 sin2x) = 0, cos2x ≠ 0
cos^2 2x - 2 sin2x = 0
sin^22x+2sin2x-1 = 0
Let y = sin2x, |y| ≤1
y^2+2y-1 = 0
y = -1+√2
sin2x = -1+√2
x1 = (1/2)[arcsin( -1+√2)] = .2135
x2 = x1+pi = 3.355
x3 = pi/2 - x1 = 1.357
x4 = pi+x3 = 4.499
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Reason:
The period of sin2x is pi.

Answer 3

cos2x=2tan2x

cos2x = 2sin2x/cos2x

2cos²2x = 2sin2x

2(1 -sin²2x) = 2six2x.....using trig formula

2 - 2sin²2x = 2sin2x....... expanding bracket

1 - sin²2x = sin2x .......... dividing by 2

sin²2x + sin2x - 1 = 0..... a quadratic in sin2x

sin2x = [-1 ± √(1 + 4)] / 2...using the quadratic formula

You should be able to finish it now

Answer 4

cos2x=2tan2x

cos2x=2(sin2x/cos2x)
cos^2(2x) = 2sin2x

1 - sin^2(2x) - 2 sin2x = 0

sin^2(2x) + 2sin(2x) - 1 = 0

let y=sin(2x)

y^2 + 2y - 1 = 0

Use the quadratic formula for roots.
Substitute back sin 2x.

Answer 5

Cos 2x=2 sin2x/cos2x;
cos^2 2x=2 sin 2x;
1- 2sin^2 2x=2 sin 2x;
Let sin 2x=2t
1- 2t^2=2 t ;
Solve the quadratic ;
you will get two values of sin2x;
sin2x= (-1+sqrt3)/2;
or,
sin 2x=(-1-sqrt3)/2;
Hence find x

Answer 6

cos 2x = 2.sin2x / cos 2x
cos² 2x = 2 sin 2x
1 - sin² 2x = 2 sin x
sin² 2x + 2 sin 2x - 1 = 0
sin 2x = [- 2 ± √8 ] / 2
sin 2x = [- 2 ± 2√2] / 2
sin 2x = - 1 ± √2
sin 2x = 0.414 (accepting +ve answer)
x = 0.214, 1.357, 3.56, 4.5 radians