Trigonometric Equation?

cos(2x) = 0

cos(2x) = cos(pi/2 + pi * k)

2x = pi/2 + pi * k

2x = (pi/2) * (1 + 2k)

x = (pi/4) * (1 + 2k)

k is an integer

Another way you could have solved this:

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

tan(x)^2 = 1

tan(x) = -1 , 1

x = pi/4 , 3pi/4 , 5pi/4 , 7pi/4 , ...

x = pi/4 + pi/2 * k

x = (pi/4) * (1 + 2k)

Solution # 1

By taking the square roots of both sides >>>> sinx = (+ -)cosx

dividing both sides by cosx >>>> tanx = (+ -)1

therefore x= tan^-1(+ - 1) =(+ -) pi/4 =(+ -) 45 degrees

Solution # 2

Exactly as you did, then

2x=cos^1 (0) = pi = 90 degrees

then, x=pi/2 or 45 deg

Divide both sides by cos^2(x):

tan^2(x) = 1

Square root both sides:

tan(x) = ± 1

Solve for each case:

tan(x) = 1

x = 45°, 225°

tan(x) = -1

x = 135°, 315°

Overall:

x = 45°, 135°, 225°, 315°

As we have not been given any restrictions in the domain:

x = 45° + 90n° (where n is any integer)

sin²(x) = cos²(x)

I think that you know that formula: cos²(x) + sin²(x) = 1

cos²(x) + sin²(x) = 1 → but you know that: sin²(x) = cos²(x)

cos²(x) + cos²(x) = 1

2.cos²(x) = 1

cos²(x) = 1/2

cos(x) = ± 1/√2

cos(x) = ± (√2)/2

First case: cos(x) = (√2)/2

x = π/4

x = 2π - (π/4) = (8π/4) - (π/4) = 7π/4

Second case: cos(x) = - (√2)/2

x = π - (π/4) = (4π/4) - (π/4) = 3π/4

x = π + (π/4) = (4π/4) + (π/4) = 5π/4

→ Solution = { π/4 ; 3π/4 ; 5π/4 ; 7π/4 }

sin²x = cos²x

divide both sides by cos²x

(sin²x)/(cos²x) = (cos²x)/(cos²x)

(sin²x)/(cos²x) = 1

note that (sin²x)/(cos²x) = (tan²x), thus

(tan²x) = 1

take the square root of both sides

√(tan²x) = √1

(tan x) = 1

x = (arctan 1)

x = 45 degrees