If 2 sin (3x -3)° = tan 240°, then the value of x in degree is:

Correct option is A

​Given:

​$$2 \sin(3x - 3)^\circ = \tan(240^\circ)$$ 

the value of x in degree  = ? 

Concept Used: 

​$$\tan(60^\circ) = \sqrt{3}$$ 

tan(180° +θ) = tan(θ)

Solution: 

​$$2 \sin(3x - 3)^\circ = \tan(240^\circ)$$ 

First, let's find the value of tan(240°). 

​$$\tan(240^\circ) = \tan(180^\circ + 60^\circ) = \tan(60^\circ)$$ 

Since the tangent function has a period of 180°, we can use the identity tan(180°  + θ) = tan(θ), but it changes sign in the third quadrant (where 240° lies). 

​$$\tan(60^\circ) = \sqrt{3}$$ 

In the third quadrant, tangent is positive, so:

$$\tan(240^\circ) = \sqrt{3}$$ ​

Substitute $$\tan(240^\circ) = \sqrt{3}$$​ into the original equation: 

​$$2 \sin(3x - 3)^\circ = \sqrt{3}$$ 

$$\sin(3x - 3)^\circ = \frac{\sqrt{3}}{2}$$ 

we know $$\sin(60)^\circ = \frac{\sqrt{3}}{2}$$ So

​$$\sin(3x - 3)^\circ = \sin(60)^\circ$$ 

$$(3x - 3)^\circ= 60^\circ$$   

​$$3x^\circ = 63^\circ$$ 

$$x = 21^\circ$$ 

Thus the value of x is 21°