trig find all solutions on the interval equation

Question

find all solutions of each equation on the interval x is > or = to 0 and < 2 pi

(tan squared x)(sec squared x)+2 sec squared x-tan squared x=2

tan2xsec2x+2sec2x-tan2x=2

Answer 1

tan = sin/cos and sec = 1/cos

sin(2x)/(cos(2x))^2 + 2/ cos(2x) - sin(2x)/cos(2x) = 2

multiplies both sides by [cos(2x)] ^2

sin(2x) + 2*cos(2x) - sin(2x)*cos(2x) = 2* [cos(2x)]^2

sin(2x) + cos(2x) [ 2 - sin(2x)] = 2 * [cos(2x)] ^2

But sin(2x) = 2 sinx cosx

2 * sinx cosx + cos(2x)[ 2 - 2 sinx cosx) = 2 [ cos2x)]^2

changes to sines and cosines : cos2x = 2*cos^2 - 1 angle argument x omitted to facilitate the notation!!!!

2 * sin * cos + [ 2*cos^2-1][ 2 - 2 sin cos] = 2[ 2 * cos^2 - 1]^2

'FOILS both sides

2 sin cos + 4 cos^2 - 4 sin cos^3 - 2 + 2 sin cos = 2[ 4 cos^4 - 4cos ^2 + 1]

Divides both sides by 2:

sin cos + 2cos^2 - 2 sin cos^3 - 1 + sin cos = 4 cos^4 - 4cos^2 + 1

combines like terms left side:

2 cos^2 - 2 sin cos^3 - 1 + 2 sin cos = 4 cos^4 - 4 cos^2 + 1

factors out 2 sin of left side

2 cos^2 - 2sin ( cos^2 + cos) = 4 cos^4 - 4 cos^2 + 1

substitutes sin = sqrt(1 - cos^2):

2 cos^2 - 2 sqrt( 1 - cos^2) (cos^2 + cos) = 4 cos^4 - 4 cos^2 + 1

Let M = cos

2M^2 - 2sqrt( 1 - M^2)( M^2 + M) = 4M^4 - 4M^2 + 1

moves polynomial from right to left, and square root from left to right

6m^2 - 4M^4 - 1 = 2sqrt(1-M^2) (M^2+M)

Divides both sides by m^2+m:

(6m^2 - 4m^4-1)/(m^2+m) = 2*sqrt(1 - M^2)

square both sides:

[(6m^2 - 4m^4-1)/(m^2+m) ]^2 = 4 (1 - M^2)

4(1-M^2)(m^2 +m)^2 - (6m^2 - 4m^4-1) = 0

4(1-M^2)(M^4 + 2m^2 + m^2) - (6m^2 - 4m^4-1) = 0

4( M^4 + 2m^2 + m^2 - m^6 - 2m^4 - m^4) - (6m^2 - 4m^4-1) = 0

4m^4 + 8m^2 + 4m^2 - 4m^6 - 8m^4 - 4m^4 - (6m^2 - 4m^4-1) = 0

4m^4 + 8m^2 + 4m^2 - 4m^6 - 8m^4 - 4m^4 - 6m^2 + 4m^4 +1 = 0

-4m^6 - 4m^4 - 2m^2 +1 = 0

4m^6 + 4m^4 + 2m^2 -1 = 0

2m^2( 4m^4 + 2m^2 + 1) = 0

2m^2 [ (2m^2)^2 + (2m^2) + 1] = 0

2m^2 [ T^2 + T + 1] = 0 where T = 2m^2

T = [-1 +or- sqrt( 1 - 4)] / 2 = [ -1 +or- sqrt(-3)]/2 = [ -1 +or- i * sqrt(3)]/2

So no good solutions come of that part.

next we have 2m^2 = 0

m^2 = 0

cos^2 = 0

cos = 0

cosine is zero at pi/2 and 3 * pi.2