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

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

asked Jan 16, 2013

+1 vote

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)

answered Jan 19, 2013