4cos2x-2sinx=1 Can anyone help me to solve this?

Answer 1

4( (1 -sinˇ2(x) ) -sinˇ2(x) ) - 2 sin x = 1

(sinx = t)

8tˇ2 + 2t - 3 =0 -> t=-3/4 or t=1/2

x= arcsin(-3/4) or x= arcsin(1/2)= TT/6 + 2kTT or
x= arcsin(1/2)= 5TT/6 + 2kTT

Answer 2

Use this trigonometric identity:
cos (2x) = 1 - 2sin^2(x)

Proceeding,
4cos(2x) - 2sinx = 1 (Given)
4*[1-2sin^2(x) ] - 2sinx = 1

To make it easier, let a = sinx, for the moment,
4* ( 1 -2a^2 ) - 2a = 1

Expanding and rearranging, we come up with the quadratic equation
8a^2 + 2a - 3 = 0

The 2 roots are 1/2 and -3/4

Therefore
sinx = 1/2 or sinx = -3/4

When sinx = 1/2, we can have 2 answers:
x = arcsin(1/2) = 30 degrees or 150 degrees

When sinx = -3/4, we can also have 2 answers:
x = arcsin(-3/4) = -48.59 degrees approximately
or x = 228.59 degrees approximately

We have 4 answers then, excluding those which we can obtain by adding or subtracting 2pie to each of these 4.

Answer 3

x=30degree
we know that: Cos2x=1-2sin(2)x ===>2(2-4sin(2)x)=1+2sinx ====>4-8sin(2)x=1+2sinx ===>8sin(2)x+2sinx-3=0 ====>sinx=1/2 ===> x=30degree

Answer 4

4(1 - 2 sin ² x) - 2 sin x - 1 = 0
4 - 8 sin ² x - 2 sin x - 1 = 0
8 sin ² x + 2 sin x - 3 = 0
(4 sin x + 3)(2 sin x - 1) = 0
sin x = - 3 / 4 , sin x = 1/2
x = 30° , 150° , 228° , 311° (to nearest degree)