Enter an exact answer and not decimal approx.

Question

Note: Enter an exact answer and not a decimal approximation.

If there is no answer type "DNE".

6tan^−1⁡(6x)=32π:

x= 

−2sin^−1⁡(8x)=2/3π:

x= 

−6sin^−1⁡(4x)=3/2π:

x= 

−2tan^−1⁡(−6x)=−4π:

x= 

Answer 1

6tan^−1⁡(6x) = 32π

tan^−1⁡(6x) = (32π)/6

tan^−1⁡(6x) = (16π)/3

But tan^-1(x) is a function with a range of (-π/2, π/2) so it cannot ever be 16π/3 therefore the solution DNE.

−2sin⁻¹⁡(8x) = (2/3)π

sin⁻¹⁡(8x) = -(1/3)π

Let θ = 8x then:

sin⁻¹⁡(θ) = -π/3

Since sin⁻¹⁡(θ) is defined for θ on (-1, 1) for ranges (-π/2, π/2), this must be in Q4

sin(sin⁻¹⁡(θ)) = sin(-π/3)

θ = -√3/2

The back-substitute to get:

8x = -√3/2

x = -√3/16

−6sin⁻¹⁡(4x) = (3/2)π

sin⁻¹⁡(4x) = -(1/4)π

sin(sin⁻¹⁡(4x)) = sin(-π/4)

4x = -√2/2

x = -√2/8

-2tan^-1(-6x) = -4π

tan^-1(-6x) = 2π

But again, tan^-1(x) is a function with a range of (-π/2, π/2) so it cannot ever be 2π therefore the solution DNE.

Answer 2

tan^-1(6x)=16π/3 same as 4π/3,,,x coordinate on unit circle is -1/2,,,y is (-√3)/2

6x=√3 taking tan of each side

x=(√3)/6

-2sin^-1(8x)=(2/3)π

on unit circle for -π/3, y=(-√3)/2,,,,,r=1

8x=(-√3)/2 taking sin of each side

sin^-1(4x)=-π/4

on unit circle for -π/4 y=(-√2)/2,,,r=1

4x=(-√2)/2

tan^-1(-6x)=2π

for 2π on unit circle x coordinate is 1 and y=0

-6x=0