SOLUTION: 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
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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): You can put this solution on YOUR website!
We don't know what the student's error was, because we don't know HOW he got his answer.
Both of the responses you have received show some poor algebraic practices; in both responses, both sides of an equation are divided by cos(x), but cos(x) might be 0.
Notice there I multiplied both sides of the equation by cos(x); since cos(x) can be 0, I will have to check any solutions I come up with.
or
We know cos(x) can't be 0, so the only possibility is that sin(x)=1. But when sin(x)=1, cos(x)=0.
So there are no solutions to the equation in ANY interval.
Correct ANSWER: No solutions
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