find solution for trig equation

Question

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

So I did this:

tan x sec² x = 2 tan x

tan x sec² x - 2 tan x = 0

tan (x) (sec² (x) - 2) = 0

tan x = 0 or sec² x - 2 = 0

x = arctan 0 sec² x = 2

x = 0 x = sec ± √2

x = π x = sec + √2 x = π/4 x = 7π/4

x = sec - √2 x = 3π/4 x = 5π/4

But my question is how can i get the values for the intervals [-π, 3π/2] 

Answer 1

The only values for x are the ones that satisfy the interval, [- π, 3π/2] and the only ones that fit are -π and 5π/4 (quadrant III).

I did it slightly different and used the identity sec²x = 1 + tan²x

tan(x)sec²(x) = 2tan(x)

tan(x)sec²(x) - 2tan(x) = 0

tan(x)(sec²(x)-2) = 0

tan(x)(1 + tan²(x) - 2) = 0

tan(x)(tan²(x) - 1) = 0

tan(x) = 0 or tan²(x) = 1

for tan(x), x = 0, π, -π, 2π, -2π, etc (multiples of π)

for tan²(x) = 1, tan(x) = 1, x = π/4, 5π/4, 9π/4, 13π/4, -3π/4, etc (multiples of 4π/4)

The only values for x within the range [-π, 3π/2] are x = -π and x = 5π/4

Answer 2

the period of tangent is pi so tan (-pi) also,is 0

The period of secant is 2pi so just subtract 2pi from each of your answers 5pi/4 - 2pi= -3pi/4

pi/4-2pi= -7pi/4 but this does not fall in the given domain

Answer 3

Just recall the period for the functions Tan and Sec...

The period for tan(x) is pi, and since tan(0) = 0, so will tan(pi) and tan(-pi).

So for the interval [-π, 3π/2] the places where tan(x) = 0 are -π, 0, and π

Also, recall that cos(x) and hence sec(x) is an even function. You already figured out positive angles that make your equation work...simply include the negatives of those angles.

Note that 7π/4 is not in the interval [-π, 3π/2] because 7π/4 > 3π/2.

Also note that -7π/4 and -5π/4 will not be in your interval as well as they are both less than -π.

Putting this all together gives the list:

-π, 0, π, -π/4, -3π/4, π/4, 3π/4, 5π/4

If you prefer to have them in order then:

-π, -3π/4, -π/4, 0, π/4, 3π/4, π, and 5π/4

Good luck, hope this was helpful for you.

Answer 4

Agreed, tan = 0

so the angle is either zer0 or 180 = pi

which of these are in the interval [-pi, 3*pi/2]???

why both of them of course...so there are 2 solutions

thus far

sec^2 = 2

cos^2 = 1/2

cos = +or- sqrt(1/2)

= +or- sqrt(2)/2

So the BASE angle where cosine is sqrt(2)/2 is

45 degrees...

starting from -pi=-180, 45 past that is -135

then -45

then 45

then 135

then 225

so there are 7 solutions