Differential Equations
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Solve the equation for solutions in the interval 0
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I need to find the exact solutions on the interval [0,2pi) for: 2sin^2(x/2)  3sin(x/2) + 1 = 0 I would start: (2sin(x/2)1)(sin(x/2)1) = 0 sin(x/2)=1/2 and sin(x/2)=1 what's next? Ok, what angle has a sin equal to say 1/2 sin
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expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by b and using that cos(b)= cos(b) sin(b)= sin(b) gives: sin(ab) = sin(a)cos(b)  cos(a)sin(b)
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Consider the function f(x)=sin(1/x) Find a sequence of xvalues that approach 0 such that (1) sin (1/x)=0 {Hint: Use the fact that sin(pi) = sin(2pi)=sin(3pi)=...=sin(npi)=0} (2) sin (1/x)=1 {Hint: Use the fact that sin(npi)/2)=1
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these must be written as a single trig expression, in the form sin ax or cos bx. a)2 sin 4x cos4x b)2 cos^2 3x1 c)12 sin^2 4x I need to learn this!! if you can show me the steps and solve it so I can learn I'd be grateful!!! 1)
asked by jim on November 1, 2006 
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In obtuse triangle PQR, P=51 degrees, p= 10cm, and the longest side, q=12cm.Draw the triangle and solve for Q to the nearest degree. I did, 10/sin 51=12/sin Q 10(sin Q)/10=12(sin 51)/10 Q= 2nd function sin 0.9325751 Q=68 degrees
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