cos^2x-sin^2x=0?

cos²(x) - sin²(x) = 0

sin²(x) = cos²(x)

sin²(x)/cos²(x) = 1

[sin(x)/cos(x)]² = 1

tan²(x) = 1

tan(x) = ± 1

First: tan(x) = 1

x = (π/4) + kπ ← where k is an integer

Second case: tan(x) = - 1

x = - (π/4) + kπ ← where k is an integer

x = 2π - (π/4) + kπ

x = (7π/4) + kπ

You can generalize writing:

x = (π/4) + k.(π/2) ← where k is an integer

cos 2x = 0

2x = 90° , 270° , 450° , 630°

x = 45° , 135 ° , 225° , 315°

cos^2x - sin^2x = 0

(cos(x) - sin(x)) (sin(x) + cos(x)) = 0

Solution:

x = (π n)/2 - π/4, n element Z

cos²x - sin²x = 0

cos²x = sin²x

sin²x / cos²x = 0

tan²x = 0

tanx = 0

x = tan⁻¹(0)

now the general solution of tanx = 0 is {nπ, n Є z}

where n = 0, 1, 2, ………

cos²x - sin²x = 0.

cos²(x) - sin²(x) = 0

sin²(x) = cos²(x)

tan²(x) = 1

tan(x) = ±1

x = (1 + 2k)π/4, where k is any integer

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

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

(1 + cos (2x))/2 - (1 - cos (2x))/2 = 0

cos (2x) = 0

2x = π/2 ± 2nπ .... n is integer

so

x = π/4 ± nπ

Cos^2x-sin^2x=0

Cos^2x=sin^2x

or

Cos^2x+sin^2x=1

add

2*Cos^2x=1

Cos^2x=1/2

Cos^2x-sin^2x = cos 2x

so cos 2x = 0

so 2x = pi/2 or 3pi/2

2x = 2npi + pi/2 and 2npi + 3pi/2

x = npi + pi/4 and npi + 3pi/4 ,,,,,,,,, where n is integer