Question 1201169: 3. 8. 1 Trigonometric Identities and Application Help me quickly as possible
A student solved the equation sin2x/cos x = 2, 0 ≤ x ≤ pi, and got pi/2. What was the student's error?
Please give the correct answer. Describe the student's error with a full explanation and show your work. Please answer me as quickly as possible.
Found 3 solutions by mananth, Theo, greenestamps:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website!
sin2x/cos x = 2,
sin 2x = 2 sinx.cosx
2sinx.cosx/cosx = 2
2sinx = 2
sin x =1
x=pi/2
But in the equation
sin2x/cos x = 2,
If x = pi/2 , cos(pi/2) =0
sin2x/0
So equation is indeterminate
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! sin2x / cosx = 2
multiply both sides of the equation by cosx to get:
sin2x = 2cosx
sin2x = 2sinxcosx, therefore:
2sinxcosx = 2cosx
divide both sidee of the equation by 2cosx to get:
sinx = 1
this occures when x = 90 degrees or 270 degrees.
90 degrees * pi/180 = pi/2
270 degrees * pi/180 = 3pi/2
therefore, this occurs when x = 90 or 270 degrees, and it occurs when x = pi/2 or 3pi/2.
here's a graph that shows when y = sin(2x) and when y = 2cos(x).
the interesection is when sin(2x) = 2cos(x)
Answer by greenestamps(13203)  (Show Source):
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