find solution for trig equation

Question

Determine solutions for tan x sec² x = 2 tan x in the interval [-π, 3π/2]    

hint – The solution for this question has to consider which quadrant the angle(s) will be in based on rotations in two directions. Examine the restrictions and the equation carefully to determine all angles that will satisfy this equation. 

I keep getting different answer with my classmates, i got π/4 and -π/4 and my friend got π and 5π/4 so i was wondering if someone can help

Answer 1

Multiplies both sides by cos^2:

tan x = 2 sin cos

Then sin/cos = 2 sin cos

sin = 2 sin cos^2

sin = 2 sin (1 - sin^2)

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

0 = 2 sin - 2 sin^3 - sin

0 = sin - 2 sin^3

0 = 2 sin^3 - sin

0 = sin ( 2 sin^2-1 )

sin = 0 or 2 sin^2 - 1 = 0

sin = 0 or sin ^2 = 1/2

sin = 0 or sin = +or- sqrt(2)/2

sin = 0:

so the BASE angle in degrees is 0 and 180

which, in the given domain is: -pi, 0, pi

sin = +or- sqrt(2)/2:

the BASE angle in degrees is 45;

under this domain:

-135, -45,45,135,215

========================================================

tan sec^2 = 2 tan

tan sec^2 - 2 tan = 0

tan ( sec^2 -2) = 0

tan = 0 or sec^2 = 2

tan=0 or cos^2 = 1/2

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

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

sin = 0:

so the BASE angle in degrees is 0 and 180

which, in the given domain is: -pi, 0, pi

cos = +or- sqrt(2)/2:

the BASE angle in degrees is 45;

under this domain:

-135, -45,45,135,215

Graph the function

y =tan(x)*sec(x)^2-2tan(x)

at desmos dot com and you will see these solution

in this interval

Answer 2

For the interval [-π, 3π/2], would put the angle in Q III if you are going counter-clockwise. If you go clockwise, then it would be Q II, Q I and Q IV. Normally angles are measured counter-clockwise.

tan(x)sec²(x) = 2tan(x)

Dividing out tan(x):

sec²(x) = 2

1/cos²(x) = 2

cos²(x) = 1/2

cos(x) = ±√2/2

If this is in Q III then x = cos^-1(-√2/2) = 5π/4 + 2πn from the Trig wheel

You can also look at it another way

tan(x)(2-sec²) = 0

which would also mean tan(x) = 0 or x = π, and sec²(x) = 2 again.

Answer 3

You both have 2 valid solutions on the interval. As with many trig equations, there are several solutions, even in this limited interval.

To find all of them, bring the 2tanx to the left-hand side of the equation, factor, and set each factor = 0:

tanxsec²x - 2tanx = 0

tanx(sec²x - 2) = 0

tanx = 0 or sec²x - 2 = 0

x = - π , 0 , π (all the multiples of π in the given interval) or secx = ±√2, so cosx = ±√2/2

x = -3π/4 , - π/4 , π/4 , 3π/4 , 5π/4

These are the 8 valid solutions on the given interval.

You can verify this by graphing the function on the left-hand side and the function on the right and looking for points of intersection. Alternatively, graph the difference function tanxsec²x - 2tanx and look for x-intercepts on the given interval.