Question 659149: I think this is trignonmetry i'm not completely sure. Could Someone help me by explaining this and how to answer it.
(a)Use a trigonometric identity and then algebra to solve the equation sin^2 x = cos x, 0 ≤ x ≤ 2π. Give all your solutions in radians.
(b)On the same set of axes draw the two graphs y = sin^2x and y = cos x for the domain [0,2π ] . Show on your graph where your solutions to part (a) are.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website!
(a)Use a trigonometric identity and then algebra to solve the equation sin^2 x = cos x, 0 ≤ x ≤ 2π. Give all your solutions in radians.
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sin^2x=cosx
Identity:sin^2x+cos^2x=1
sin^2x=1-cos^2x
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1-cos^2x=cosx
cos^2x+cosx-1=0
let u=cosx
u^2+u-1=0
solve for u by following quadratic formula:
a=1, b=1, c=-1
ans
u≈-1.62 (reject, not in domain)
or
u≈.618
cosx=.618
from calculator:
x=.905 and 5.378 radians (in quadrants I and IV where cos>0)
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b)On the same set of axes draw the two graphs y = sin^2x and y = cos x for the domain [0,2π ] . Show on your graph where your solutions to part (a) are.
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I don't have the means to draw graphs for you, but I will provide the coordinates with which you can draw the graphs yourself.
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For cosx for 1 period (0-2π)
(0,1), (π/2,0), (π,-1), (3π/2,0), (2π,1)
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For sin^2x for 1 period
(0,0), (π/2,1), (π,0), (3π/2,1), (2π/0)
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If you draw these two curves correctly, you will find that they intersect at coordinates (0.905,0.618) and (5.378,0.618)
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