Find solutions of the trigonometric equation sin 12x+cos 6x=0?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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