Sin2x=1/√3cos2x ............

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To solve this equation, we can use the trigonometric identity sin(2x) = 2sin(x)cos(x) and cos(2x) = 2cos^2(x) - 1.

Given that sin(2x) = 1/√3cos(2x), we can substitute these identities into the equation:

2sin(x)cos(x) = 1/√3(2cos^2(x) - 1)

Expand and simplify:

2sin(x)cos(x) = 2cos^2(x)/√3 - 1/√3

Now, divide both sides by 2cos(x) to isolate sin(x):

sin(x) = cos(x)/√3 - 1/(2√3cos(x))

We know that sin(x)/cos(x) = tan(x), so we can rewrite sin(x) and cos(x) in terms of tan(x):

tan(x) = 1/√3 - 1/(2√3tan(x))

To solve for tan(x), multiply both sides by 2√3tan(x):

2√3tan^2(x) = 2 - √3tan(x)

Rearrange the equation to set it to zero:

2√3tan^2(x) + √3tan(x) - 2 = 0

Now, this is a quadratic equation in terms of tan(x). To find the values of tan(x), you can use the quadratic formula:

tan(x) = [-√3 ± √(3 + 24√3)] / 4√3

Therefore, the solutions for the equation sin(2x) = 1/√3cos(2x) are the values for x that satisfy the equation tan(x) = [-√3 ± √(3 + 24√3)] / 4√3.