What is the exact answer to cos^2x - sin^2x = 0 for between 0 and 2pi?
I got x = pi/4 and x = 5pi/4. But the answer key says there are two additional answers which are 3pi/4 and 7pi/4. I don't know how to get these. Any help is much appreciated!
cos^2x - sin^2x = 0
cos (2x) = 0
2x = π/2 or 2x = 3π/2
x = π/4 or π = 3π/4
but the period of cos (2x) is π
so by adding π to any answer will yield a new answer
new answers :
π/4+π = 5π/4
3π/4 + π = 7π/4
Since I converted to cos 2x, there would be 2 complete cosine curves for 0 to 2π
other way:
cos^2 x = sin^2 x
sin^2 x/cos^2 x = 1
tan ^2 x = 1
tan x = ± 1
x = 45°, 135° , 225° , 315°
or
π/4 , 3π/4 , 5π/4 , 7π/4
Thanks!
There is an expression some people use that says, “What you put into it is what you get out of it.” People might use this expression to describe your skills at a sport or activity and how that relates to the amount of time and effort you spend practicing that activity. Does this expression apply to functions? How?
There is an expression some people use that says, “What you put into it is what you get out of it.” People might use this expression to describe your skills at a sport or activity and how that relates to the amount of time and effort you spend practicing that activity. Does this expression apply to functions? How? Give an example to support your answer.
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To find all the solutions to the equation cos^2x - sin^2x = 0 between 0 and 2pi, we can start by using the trigonometric identity cos^2x - sin^2x = cos(2x). Therefore, we need to solve cos(2x) = 0.
The solutions to cos(2x) = 0 lie at the x-values where the cosine function equals zero. This occurs at two specific angles: 2x = pi/2 and 2x = 3pi/2.
To find all the solutions in the range of 0 to 2pi, we divide each angle by 2:
2x = pi/2 => x = pi/4
2x = 3pi/2 => x = 3pi/4
So far, we have found x = pi/4 and x = 3pi/4 as solutions to the equation.
However, there are two additional solutions that need to be considered. Since cosine is a periodic function with a period of 2pi, we can add or subtract any multiple of 2pi to the solutions we have already found.
Adding 2pi to pi/4 gives us: x = pi/4 + 2pi = 9pi/4
Adding 2pi to 3pi/4 gives us: x = 3pi/4 + 2pi = 11pi/4
So, the complete set of solutions to cos^2x - sin^2x = 0 between 0 and 2pi is:
x = pi/4, 3pi/4, 9pi/4, 11pi/4.