Question

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

Answer 1

Answer:

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

Step-by-step 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 $\cos{2x}$:

$\dfrac{\sin{2x}}{\cos{2x}}=4$

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

$\dfrac{\sin{2x}}{\cos{2x}}=\tan{2x}$

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

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