Find all solutions in the interval [0, 2π). (6 points)

sin^2x - cos^2x = 0

Question

Find all solutions in the interval [0, 2π). (6 points) 

sin^2x - cos^2x = 0

mathstudent55 · Answer

Treat this as a quadratic equation, a difference of squares.
Factor it, then set each factor equal to zero and solve.

naveenbhatia1312 · Answer

would the final answer be pi/4 or are there more or different solutions. @mathstudent55

mathstudent55 · Answer

pi/4 is a correct solution, but there are more solutions in that interval.

mathstudent55 · Answer

Did you factor the equation?

naveenbhatia1312 · Answer

pi/4, 3pi/4, 5pi/4, and 7pi/4?

mathstudent55 · Answer

Correct.

mathstudent55 · Answer

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naveenbhatia1312 · Answer

Find the exact value by using a half-angle identity. 
cos(-pi/8)

mathstudent55 · Answer

$\sin^2 x - \cos^2 x = 0$

$(\sin x + \cos x)(\sin x - \cos x) = 0$

$\sin x = -\cos x$   or   $\sin x = \cos x$

x = pi/4, 5pi/4    or   x = 3pi/4, 7pi/4

mathstudent55 · Answer

What is the half-angle identity for cos?

naveenbhatia1312 · Answer

?

mathstudent55 · Answer

Your second problem is asking you to use the half-angle identity to find the cos(-pi/8)
You must start with the half-angle identity for cosine.
What is it?

mathstudent55 · Answer

$\cos \dfrac{\theta}{2} = \pm \sqrt{\dfrac{1 + \cos \theta}{2}     }$

naveenbhatia1312 · Answer

ohh i didnt know that

mathstudent55 · Answer

pi/8 = (pi/4) / 2
Use theta = pi/4 since you know the exact value of cos pi/4
Substitute the values into the half-angle identity above.
Then you need to choose the sign since the identity has $\pm$ in it.
To choose the sign, figure out what quadrant -pi/8 is in and what the sign of the cosine is in that quadrant.

mathstudent55 · Answer

sorry, but gtg