SOLUTION: I am lost with these questions, please any help would be amazing! 1)Solve 2(sin^2)x-1 =0 2)Solve 4sin2x - 4 = 0 in [0,2pi) 3)Solve cos(x)-1 = sin(x) in [0,2pi) Thank you so

Question 466595: I am lost with these questions, please any help would be amazing!
1)Solve 2(sin^2)x-1 =0
2)Solve 4sin2x - 4 = 0 in [0,2pi)
3)Solve cos(x)-1 = sin(x) in [0,2pi)
Thank you so much!!
- Michelle
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1)Solve 2(sin^2)x-1 =0
2sin^2(x) = 1
sin^2(x) = 1/2
sin(x) = sqrt(2)/2
---
x = pi/4 or x = (3/4)pi
-------------------------------
2)Solve 4sin2x - 4 = 0 in [0,2pi)
sin(2x) = 1
2x = pi/2
x = pi/4
-------------------------------
3)Solve cos(x)-1 = sin(x) in [0,2pi)
cos(x)-sin(x) = 1
---
Square both sides:
cos^2-2sin*cos+sin^2 = 1
-2sin(x)cos(x) + 1 = 1
-2sin(x)cos(x) = 0
-sin(2x) = 0
2x = 0 or 2x = pi
x = 0 or x = pi/2
=======================
Check for extraneous answers:
x = 0 ??
cos(0)-1 = sin(0)
1-1 = 0
0 = 0
===============
x = pi/2
cos(x)-1 = sin(x)
cos(pi/2)-1 = sin(pi/2)
0-1 = 1
-1 = 1
No
==================
Only solution:
x = 0
==============================
Cheers,
Stan H.
================
Cheers,
Stan H.

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