SOLUTION: Solve:
{{{sin2x=cosx}}}, {{{-2pi<=x<=pi}}}
Algebra.Com
Question 553053: Solve:
,
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Solve:
,
-----------------
2sinx*cosx = cosx
2sinx*cosx - cosx 0
cos(x)*(2sin(x) - 1) = 0
cos(x) = 0
x = pi/2, -pi/2, -3pi/2
------
2sin(x) = 1
sin(x) = 1/2
x = -11pi/6, -7pi/6, pi/6, 5pi/6
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Since \ =\ 2\sin(x)cos(x))
we can write:
cos(x)\ =\ cos(x))
Multiply by })
\ =\ 1)
\ =\ \frac{1}{2})
)
Use the unit circle, recalling that sin of the angle is the 
-coordinate of the point of intersection of the terminal ray with the unit circle and find all angles where \ =\ \frac{1}{2})
in your given interval. Note that the given interval is one and a half trips around the circle. Hint: start at 
and go backwards.
John
My calculator said it, I believe it, that settles it
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