Question

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

sin^2x - cos^2x = 0

Answer 1

Find all solutions in the interval [0, 2pi] for sin^2x-cos^2x=0
C) x= pi/4 and 3pi/4 and 5pi/4 and 7pi/4
- find the places where the cosx = +/- sinx; these occur at the odd pi/4

and do you maybe know how to do this?????
The expression: secx+cscx/tanx+cotx is to be the left hand side of an equation the is an identity. Which one of the following four expressions can be used as the right hand side of the equation to complete the identity???
A)sinx+cosx
- convert sec, csc, cot to sin & cos to get (s=sin, c=cos)
(1/c + 1/s) / (s/c + c/s) =
(s+c)/sc / (s^2 + c^2)/sc =
s+c / 1

and maybe you can do this one????
Find all the solutions to the equation in the interval [0, 2pi] of cosx=sin2x
A) pi/6 and pi/2 and 5pi-6, and 3pi/2
- as in the first, you need to learn your values of sin, cos for the pi/6, pi/4, & pi/2 values. KNOW them

Answer 2

what is the answer?

Answer 3

sin(2x) = sin(4x)
sin(2x) = sin(2*2x)
sin(2x) = 2sin(2x)cos(2x)
2sin(2x)cos(2x) - sin(2x) = 0
sin(2x) * (2cos(2x) - 1) = 0
sin(2x) = 0 or cos(2x) = 1/2
x = 0 + n•π/2, π/6 + n•π, 5π/6 + n•π

x = 0, π/6, π/2, 5π/6, π, 7π/6, 3π/2, 11π/6
help better