Question

Deremine the general solution of sin 2x - 4cos2x​

Answer 1

`x=(1)/(2)\left(\tan^(-1){4}+\pi k\right)`, where k is any integer

Explanation:

Let's start by getting all the trig functions on one side and all the constants on the other. We can do this by dividing both sides by
`cos(2x)`:

`(sin(2x))/(cos(2x))=4`

This ratio looks familiar! It just so happens that the tangent function is defined as the ratio of sine to cosine. In our case:

`(sin(2x))/(cos(2x))=tan(2x)`

Substituting this back into our equation, we have
`tan(2x)=4`. We can unwrap the 2x by applying the inverse tangent function to both sides, giving us
`2x=\tan^(-1){4`. Note, this specific solution only accounts for values of 2x between 0 and 2π radians. To make it general, we can add the term πk to the end of the right side, where k is any integer. We use π as a coefficient because the tangent function has a period of π radians, and it repeats its values every period.

Finally, we divide both sides of the equation by 2 to isolate x, giving us

`2x=\tan^(-1){4}+\pi k\\x=(1)/(2)\left(\tan^(-1){4}+\pi k\right)`