How do I solve this Trigonometric Identities? step by step?

Question

6 tan^2 x - 4 sin^2x = 1 for 0 < or = to x < or = 2pi

Answer 1

6tan²x - 4sin²x = 1 for [ 0 < x < 2π ]

Note : [ TanA = sinA / cosA ]

6[(sin²x) / (cos²x)] - 4sin²x = 1

Multiply each side by cos²x

6[(sin²x) / (cos²x)] (cos²x)- 4(sin²x) (cos²x)= 1(cos²x)

Simplify

6(sin²x) - 4(sin²x) (cos²x)= (cos²x)

Formula :[ sin²x + cos²x = 1 ,  cos²x = 1 - sin²x ]

6(sin²x) - 4(sin²x) (1 - sin²x)= (1 - sin²x)

6(sin²x) - 4(sin²x)  + 4(sin²x)(sin²x)= (1 - sin²x)

2(sin²x) + 4(sin⁴x) = (1 - sin²x)

Add ( sin²x - 1) to each side

2(sin²x) + 4(sin⁴x) + ( sin²x - 1) = (1 - sin²x) + ( sin²x - 1)

Simplify

4(sin⁴x) + 3 sin²x - 1 = 0

Let sin²x = t

There fore.

4t² + 3t - 1 = 0

Now solve the equation factor method

4t² + 4t - t - 1 = 0

Take out common factors.

4t(t + 1) - 1(t + 1) = 0

Take out common factors.

(4t - 1)(t + 1) = 0

4t - 1 = 0  or   t + 1 = 0

Add 1 to each side

4t - 1 + 1 = 0 + 1

4t = 1

Divide each side by 4.

4t / 4 = 1 / 4

t = 1/4

And  t + 1 = 0

Subtract 1 from each side.

t + 1 - 1 = - 1

t = - 1

There fore

t = 1 / 4     and t = - 1

But  t = sin²x

sin²x = 1 / 4  and sin²x = - 1

sin²x = 1 / 4

Then

sinx = 1 / 2

sinx = sin 30

x = 30 = π / 6

sin²x = - 1

Than

sinx = √(-1)

x = sin^-1√(-1)

There fore x = π / 6 or sin^-1√(-1)