Find solutions of the trigonometric equation sin 12x+cos 6x=0?
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We need the solutions of sin 12x + cos 6x = 0.
sin 12x + cos 6x = 0
=> 2*sin 6x*cos 6x + cos 6x = 0
let cos 6x = y
=> 2*sqrt(1 - y^2)*y + y = 0
=> y(2*sqrt(1 - y^2) + 1) = 0
=> y = 0
2*sqrt(1 - y^2) + 1 = 0
=> 4*(1 - y^2) = 1
=> 1 - y^2 = 1/4
=> y^2 = 3/4
=> y = sqrt 3/2
cos 6x = 0 and cos 6x = sqrt 3/2
6x = arc cos 0 and x = (1/6)*arc cos (sqrt 3/2)
x = pi/12 + n*pi/3 and x = pi/36 + n*pi/3
The solution of the equation is x = pi/12 + n*pi/3 and x = pi/36 + n*pi/6, where n is a positive or negative integer.
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We'll change sin 12x, using the double angle identity, into:
sin 12x = sin 2*(6x) = 2 sin 6x*cos 6x
We'll re-write the equation:
2 sin 6x*cos 6x + cos 6x = 0
We'll factorize by cos 6x:
cos 6x(2 sin 6x + 1) = 0
We'll set each factor as zero:
cos 6x = 0
6x = +/-arccos 0 + 2kpi
6x = +/-(pi/2) + 2kpi
We'll divide by 6:
x = +/-(pi/12) + kpi/3
2 sin 6x + 1 = 0
sin 6x = -1/2
6x = (-1)^k*arcsin(-1/2) + kpi
x = (-1)^(k+1)*(pi/36) + kpi/6
The solutions of trigonometric equation are: {+/-(pi/12) + kpi/3 ; k integer}U{(-1)^(k+1)*(pi/36) + kpi/6 ; k integer}.
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