Solve the equation $sin(2x) +3cos(2x) = 0$ ,for $0<x<180$ ?

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Answer 1 · 2017-11-14T17:39:04

$x ~~ = 54.2^@$ and $x ~~ 141.2^@$

Given: $sin(2x) +3cos(2x) = 0$

Subtract $3cos(2x)$ from both sides:

$sin(2x) = -3cos(2x)$

Divide both sides by $cos(2x)$

$sin(2x)/cos(2x) = -3; 0^@ < x < 180^@$

Substitute $tan(2x)$ for $sin(2x)/cos(2x)$ :

$tan(2x) = -3; 0^@ < x < 180^@$

Use the inverse tangent function on both sides:

$2x = tan^-1(-3); 0^@ < x < 180^@$

Divide both sides by 2:

$x = 1/2tan^-1(-3)$

If we were to use a calculator to evaluate the above equation, we would obtain a negative angle that would be outside of the specified domain.

We need to add the cyclical nature of the inverse tangent function that has a period of $180^@$ :

$x = 1/2{n(180^@)+tan^-1(-3)}; n in ZZ$

To obtain the two values within the domain, we evaluate the above equation with $n = 1$ and $n = 2$ :

$x = 90^@+ 1/2tan^-1(-3); 0^@ < x < 180^@$

$x ~~ = 54.2^@$

$x = 180^@ + 1/2tan^-1(-3); 0^@ < x < 180^@$

$x ~~ 141.2^@$

Solve the equation sin(2x) +3cos(2x) = 0sin(2x) +3cos(2x) = 0 ,for 0<x<1800<x<180?