Answer to Question #129076 in Trigonometry for Ali

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) = 1 - 2 sin^2(x)" . Now equation looks like

"1\u22122sin^2(x) + 0.5 sin(x) = 0"

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

"4 sin^2 (x) - sin(x) -2 =0"

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

"sin(x) = 1\/8 (1 \u00b1\\sqrt{1+32}) =1\/8 (1 \u00b1\\sqrt{33})"

For the first root ("sin x = 1\/8(1+\\sqrt{33})" ) the solution is"x = (-1)^k arcsin(1\/8(1+\\sqrt{33})) + \\pi k \u2248\n (-1)^k + \\pi k"

k =0: "x_1 = 1 rad \u2248 57\u00b0"

k = 1: "x_2 = -1 + \\pi = 2.14 rad \u2248123\u00b0"

The same for the second root:

"x = (-1)^k arcsin(1\/8(1-\\sqrt{33})) + \\pi k \u2248\n (-1)^k*(-0.635) + \\pi k"

"k =1: x_3= (0.635 + \\pi )rad\u22483.78 rad \u2248 216\u00b0"

"k =2: x_4= (-0.635 + 2\\pi) rad\u22485.645 rad \u2248 323\u00b0"


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