Thru what method would you solve this problem?
cos^2x = sin^2x +1
Linear Method
Factoring
Quadratic Formula
Squaring
Are we limited to those four methods?
If not, then I will use substitution as the first step, and then I will see the resulting equation. It could then be by factoring or by using the quadratic formula, or by whatever.
cos^2x = sin^2x +1
Rewriting that,
cos^2(x) = sin^2(x) +1 ----(i)
First, I will use the trig identity
sin^2(x) +cos^2(x) = 1
to transform Eq.(i) into sines only.
From that trig identity,
cos^2(x) = 1 -sin^2(x)
Substitute that into (i),
1 -sin^2(x) = sin^2(x) +1
Collect like terms,
-sin^2(x) -sin^2(x) = 1 -1
-2sin^2(x) = 0
Divide both sides by -2,
sin^2(x) = 0
Take the square roots of both sides,
sin(x) = 0
Take the arcsine of both sides,
x = arcsin(0) ----***
If we limit the domain of x from 0 to 360 degrees only,
x = 0, 180, or 360 degrees. ----answer.
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Another way.
We transform Eq.(i) into cosines only.
From the same trig identity we used above,
sin^2(x) = 1 -cos^2(x)
Substitute that into (i),
cos^2(x) = [1 -cos^2(x)] +1
cos^2(x) = 1 -cos^2(x) +1
cos^2(x) +cos^2(x) = 1 +1
2cos^2(x) = 2
Umm, can be by factoring.
2cos^2(x) = 2
2cos^2(x) -2 = 0
2(cos^2(x) -1) = 0
Divide both sides by 2,
cos^2(x) -1 = 0
That quadratic equation is a difference of two squares,
factoring that,
(cos(x) +1)*(cos(x) -1) = 0
cos(x) +1 = 0
cos(x) = -1
x = arccos(-1)
x = 180 degrees
cos(x) -1 = 0
cos(x) = 1
x = arccos(1)
x = 0, or 360 degrees
Therefore, if x is from 0 to 360 degrees only, then,
x = 180, 0, or 360 degrees. ---same as in the first way above.
You can also use a double angle formula :Originally Posted by jp25111
cos^2(x) - sin^2(x) = 1
cos(2x) = 1
so 2x = 0 or 2pi or 4pi ...
so x must be a multiple of pi - as in ticbol's solution.
Check whether this always gives a solution (always a good idea for trig. equations):
cos (k*pi) = +1 or -1, sin(*pi) = 0,
so these are indeed solutions of the problem.