Question 167797This question is from textbook Trigonometry: A Graphing Approach
: Find the solutions of the equation in the interval [0,2∏)
cos(X/2)-sin(X)= 0
This question is from textbook Trigonometry: A Graphing Approach
Found 2 solutions by jim_thompson5910, Alan3354:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let u=x/2. So this means that 2u=2(x/2)=x (ie 2u=x)
cos(x/2)-sin(x)= 0 ... Start with the given equation
cos(u)-= 0 ... Replace x/2 with u and x with 2u
cos(u)-2sin(u)cos(u)= 0 ... Use the identity sin(2A) = 2sin(A)cos(A)
cos(u)(1-2sin(u))= 0 ... Factor out the GCF cos(u)
cos(u)=0 ... or ... 1-2sin(u)=0 ... Set each factor equal to zero
Let's solve the first equation cos(u)=0
By taking the arccosine of both sides, we get u=pi/2 or u=3pi/2
But remember, we let u=x/2. So this means that x/2=pi/2 or x/2=3pi/2
Solving for x, we get x=pi or x=3pi (we must discard this solution as it is NOT in the given interval)
So the first solution is x=pi
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Now let's solve the second equation 1-2sin(u)=0
-2sin(u)=-1 ... Subtract 1 from both sides
sin(u)=1/2 ... Divide both sides by -2
Taking the arcsine of both sides gives us u = pi/6 or u = 5pi/6
Once again, recall that we said that u = x/2. So x/2=pi/6 or x/2=5pi/6
So this means that the values of x are x=pi/3 or x=5pi/3 (both of which are within the given interval)
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Answer:
So the three solutions in the interval [0,2∏) are
x=pi, x=pi/3 or x=5pi/3
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Find the solutions of the equation in the interval [0,2∏)
cos(X/2)-sin(X)= 0
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cos(x/2) - 2sin(x/2)cos(x/2) = 0
cos(x/2) = 0
x/2 = 90, 270
x = 180º = pi (540 is 180, and exceeds 2pi)
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1 - 2sin(x/2) = 0
sin(x/2) = 1/2
x/2 = 30º, 150º
x = 60º, 300º = pi/3, 5pi/3
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