How do I solve sec(x) * tan(x) = sqrt(2)?

Question

All I know is that it involves the quadratic formula at some point after using trig identities.

Answer 1

use a scientific calculator to find square root 2 or look it up in your root tables

Answer 2

sec(x) = 1/cos(x) (identity)

so the equation becomes:

1/cos(x) x tan(x) = √2

tan(x)/cos(x) = √2

tan(x) = sin(x)/cos(x) (identity)

sin(x)/(cos(x))^2 = √2

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

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

sin(x)/(1 - (sin(x))^2) = √2

multiply both sides by (1 - (sin(x))^2) and then rearrange and solve using the quadratic formula.

  

Answer 3

sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x), so:

sec(x) * tan(x) = sin(x)/(cos^2(x)) = sin(x)/(1-sin^2(x)) = sqrt(2)

sin(x) = sqrt(2) - sqrt(2)*sin^2(x)

Let a = sin(x). Then,

a = sqrt(2) - sqrt(2)*a^2

sqrt(2)*a^2 + a - sqrt(2) = 0

Solve this using the quadratic formula to find that:

a = sin(x) = sqrt(2)/2 or a = sin(x) = -sqrt(2).

-1 <= sin(x) <= 1, so sin(x) = -sqrt(2) has no solutions in x.

Therefore, sin(x) = sqrt(2)/2, so:

x = pi/4 + 2pi*k or x = 3pi/4 + 2pi*k (k is an integer)

Answer 4

sqrt(2) = sec(x) * tan(x)

= tan x * (1 / cos x)

= sin x / (cos x)^2

= sin x / [1 - (sin x)^2]

let sin x = y,

sqrt(2) [ 1- y^2] = y

y^2 + y/√2 - 1 = 0

y = {-1/√2  +- √[1/2 - 4*1*(-1)]} / 2*1

= {-1/√2  +- √[9/2]} / 2

= [-1 +- 3] / 2√2

since -1 ≤ y = sin x ≤ 1,

sin x = [-1 + 3] / 2√2

= 2 / 2√2

= 1 / √2

x = (4n-3)π/2 +- π/4