Solve the equation sin^2x-cos^2x=0 over the interval [0, 2pi). Having trouble getting started.
cos ^ 2 ( x ) = 1 - sin ^ 2 ( x )
sin ^ 2 ( x ) - cos ^ 2 ( x ) = 0
sin ^ 2 ( x ) - [ 1 - sin ^ 2 ( x ) ] = 0
sin ^ 2 ( x ) - 1 + sin ^ 2 ( x ) = 0
2 sin ^ 2 ( x ) - 1 = 0
2 sin ^ 2 ( x ) = 1 Divide both sides by 2
sin ^ 2 ( x ) = 1 / 2
sin ( x ) = ± 1 / sqrt ( 2 )
Solutions :
x = pi / 4
[ pi / 4 = 45 ° , sin ( 45 ° ) = 1 / sqrt ( 2 ) ]
x = 3 pi / 4
[ 3 pi / 4 = 135 ° , sin ( 135 ° ) = 1 / sqrt ( 2 ) ]
x = 5 pi / 4
[ 5 pi / 4 = 225 ° , sin ( 225 ° ) = - 1 / sqrt ( 2 ) ]
x = 7 pi / 4
[ 7 pi / 4 = 315 ° , sin ( 315 ° ) = - 1 / sqrt ( 2 ) ]
To solve the equation sin^2x - cos^2x = 0, first, let's simplify the equation using the trigonometric identity sin^2(x) - cos^2(x) = -cos(2x).
So, we have -cos(2x) = 0.
To find the solutions within the given interval of [0, 2π), we can set up two cases:
Case 1: -cos(2x) = 0
Solve for x:
cos(2x) = 0
Now, we know that cos(x) = 0 when x = π/2 or x = 3π/2.
But we have 2x, so we double those values:
x = π/4 or x = 3π/4.
Case 2: Consider the periodicity of cosine function
Since cosine is a periodic function, we need to find other solutions within the given interval by considering the periodicity of cos(2x).
The period of cos(2x) is π, so we can add π to our previous solutions:
x = π/4 + π = 5π/4 or x = 3π/4 + π = 7π/4.
But this is still not within the interval [0, 2π). So, let's subtract 2π from our solutions:
x = 5π/4 - 2π = -3π/4.
Now, we have found all solutions within the interval [0, 2π):
x = π/4, 3π/4, 5π/4, -3π/4, and 7π/4.
So, the equation sin^2x - cos^2x = 0 has 5 solutions within the interval [0, 2π): π/4, 3π/4, 5π/4, -3π/4, and 7π/4.