Question 167797
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