Question #129076

 Solve the equation for all x  0<x<360

  

Cos(2x) + ½sin(x) = 0 


1
Expert's answer
2020-08-11T19:11:31-0400

We know that cos(2x)=12sin2(x)cos(2x) = 1 - 2 sin^2(x) . Now equation looks like

12sin2(x)+0.5sin(x)=01−2sin^2(x) + 0.5 sin(x) = 0

Multiply the whole equation by -2 (just for convenience):

4sin2(x)sin(x)2=04 sin^2 (x) - sin(x) -2 =0

Now we have a standard quadratic equation for sin(x)

sin(x)=1/8(1±1+32)=1/8(1±33)sin(x) = 1/8 (1 ±\sqrt{1+32}) =1/8 (1 ±\sqrt{33})

For the first root (sinx=1/8(1+33)sin x = 1/8(1+\sqrt{33}) ) the solution isx=(1)karcsin(1/8(1+33))+πk(1)k+πkx = (-1)^k arcsin(1/8(1+\sqrt{33})) + \pi k ≈ (-1)^k + \pi k

k =0: x1=1rad57°x_1 = 1 rad ≈ 57°

k = 1: x2=1+π=2.14rad123°x_2 = -1 + \pi = 2.14 rad ≈123°

The same for the second root:

x=(1)karcsin(1/8(133))+πk(1)k(0.635)+πkx = (-1)^k arcsin(1/8(1-\sqrt{33})) + \pi k ≈ (-1)^k*(-0.635) + \pi k

k=1:x3=(0.635+π)rad3.78rad216°k =1: x_3= (0.635 + \pi )rad≈3.78 rad ≈ 216°

k=2:x4=(0.635+2π)rad5.645rad323°k =2: x_4= (-0.635 + 2\pi) rad≈5.645 rad ≈ 323°


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