# Solve the equation 2-3cosx+2square root((1-cos4x)/2)=3sinx

*print*Print*list*Cite

### 1 Answer

First, we'll move the term 3cosx to the right side:

2 + 2sqrt((1-cos4x)/2) = 3sinx + 3cosx

We'll apply the half angle identity for the term sqrt((1-cos4x)/2)= sin 2x.

2 + 2sin 2x = 3(sin x + cos x)

We'll apply the double angle identity for sin 2x:

sin 2x = 2 sin x*cos x

We'll solve this equation algebraically.

We'll move all terms to one side:

2sin2x - 3(sinx+cosx) + 2 = 0

2*2sin x*cos x - 3(sinx+cosx) + 2 = 0

We'll note sin x + cos x = y.

We'll raise to square and we'll get:

(sin x + cos x)^2 = y^2

(sin x)^2 + (cos x)^2 + 2sin x*cos x = y^2

From Pythagorean identity:

(sin x)^2 + (cos x)^2 = 1

1 + 2sin x*cos x = y^2

2sin x*cos x = y^2 - 1

We'll re-write the given equation in y:

2*(y^2 - 1) - 3(y) + 2 = 0

We'll remove the brackets:

2y^2 - 2 - 3y + 2 = 0

We'll eliminate like terms:

2y^2 - 3y = 0

We'll factorize by y:

y(2y - 3) = 0

We'll put y = 0

But y = sin x + cos x:

sin x + cos x = 0

We'll divide by cos x:

tan x + 1 = 0

tan x = -1

x = -pi/4 + kpi

2y + 3 = 0

2y = 3

y = 3/2

But the range of values of y is [-2;2].

Maximum of the sum: sin x + cos x = 1 + 1 = 2

Minimumof the sum: sin x + cos x = -1-1 = -2

We'll work with substitution:

sin x = 2t/(1+t^2)

cos t = (1-t^2)/(1+t^2)

2t/(1+t^2) + (1-t^2)/(1+t^2) = 3/2

4t + 2 - 2t^2 = 3 + 3t^2

We'll move all terms to one side:

t1 = [-4+sqrt(16-20)]/2

Since delta is negative, the equation has no real solutions.

**The equation will have only one solution: ****x = -pi/4 + kpi.**