�ã3tanx=3tanxcscx-�ã3tanx
those are supposed to be square root symbols. I'm not sure what happened...
so you want
√(3tanx) = 3tanxcscx - √(3tanx)
then
2√(3tanx) = 3 (sinx/cosx)(1/sinx))
= 3/cosx
square both sides
4(3tanx) = 9/cos^2 x
12 sinx/cosx = 9/cos^2 x
12 sinx = 9/cosx
12 sinxcosx = 9
2 sinxcosx = 9/6 = 3/2
sin (2x) = 3/2
But the sine of anything cannot > 1
your equation has no real solution.
To solve the equation 3tan(x) = 3tan(x)csc(x) - 3tan(x), we can simplify it step by step. Here's how:
Step 1: Distribute 3tan(x) on the right side of the equation:
3tan(x) = 3tan(x)csc(x) - 3tan(x)
Step 2: Remove the common factor of 3tan(x):
3tan(x) - 3tan(x) = 3tan(x)csc(x) - 3tan(x)
Step 3: Simplify the left side of the equation:
0 = 3tan(x)csc(x) - 3tan(x)
Step 4: Factor out 3tan(x) on the right side of the equation:
0 = 3tan(x)(csc(x) - 1)
Step 5: Set each factor equal to zero and solve for x:
a) 3tan(x) = 0
To solve this equation, we divide both sides by 3:
tan(x) = 0
Using a unit circle or trigonometric identities, we can find the solutions for this equation. The solutions for tan(x) = 0 are x = 0, π, 2π, 3π, etc.
b) csc(x) - 1 = 0
To solve this equation, we add 1 to both sides:
csc(x) = 1
Using trigonometric identities, we know that csc(x) = 1 when sin(x) = 1. The solutions for sin(x) = 1 are x = π/2, 5π/2, 9π/2, etc.
So the solutions to the original equation are: x = 0, π, 2π, 3π, etc. and x = π/2, 5π/2, 9π/2, etc.