Solve for x over (0,2pi): 1) tan^2 x = 2sec x - 1; 2) sin x - sqrt3 cos x = 1.?

Answer 1

First part:

tan^2(x) = 2sec(x) - 1

or sec^2(x) - 1 = 2sec(x) - 1

or sec^2(x) - 2*sec(x) = 0

or {sec(x) - 2}*sec(x) = 0

Therefore either sec(x) = 0 or sec(x) = 2
i.e. either cos(x) = 1/0 or cos(x) = 1/2.

Since division by 0 is undefined so cos(x) = 1/0 is not admissible.

Hence, cos(x) = 1/2
As 0 < x < 2*pi, so x = pi/3 or 5pi/3


Second part:

sin(x) - {sqrt(3)}*cos (x) = 1

Dividing both sides by 2

(1/2)*sin(x) - {sqrt(3)/2}*cos (x) = 1/2

or {sin(x)}*{cos(pi/3)} - {cos(X)}*{sin(pi/3)} = sin(pi/6)

or sin{x - (pi/3)} - sin(pi/6) = 0

or 2*cos{x + (pi/3)}*sin{x - (pi/2)} = 0


Either
cos{x + (pi/3)} = 0

i.e. x + (pi/3) = n(pi/2)
[where n = +/-1,+/-3,+/-5,.....]

i.e. x = pi/6, 7*pi/6
[for n = 1 and 3; for other values of 'n' x is outside the interval 0 < x < 2*pi]]

Or
x - (pi/2) = n*pi
[where n = 0, +/-1, +/-2, +/-3, .......]

i.e. x = pi/2, 3*pi/2


Answer: x = pi/6, pi/3, 3*pi/2, 7*pi/6

Answer 2

you do it