Find the exact solutions of the equations in the given interval

Question

csc³(x) + csc²(x) = 4 csc(x) + 4 in [0, 2π)

-3tan (3θ/2) = √3 in (-2π,π)

cot⁴(x) = 4 csc²(x) - 7 in [0, 2π)

Answer 1

csc³(x) + csc²(x) = 4 csc(x) + 4 Put all terms on left side.

csc³(x) + csc²(x) - 4 csc(x) - 4 = 0. Factor by grouping.

csc²(x) (csc x + 1) -4 (csc x + 1) = 0

(csc²(x) - 4 ) (csc x + 1) = 0

csc²(x) - 4 = 0 and csc x + 1= 0

csc²(x) = 4 and csc x = - 1 Remembering that Csc x = 1/Sin x

1 / sin² x = 4 and 1 / sin x = -1

So sin² x = 1 /4 and sin x = -1 Take square root of first eon.

sin x = ± 1 /2 and sin x = -1 From the unit circle

x = π / 6, 5π / 6, 7π / 6, 11π / 6. and x = 3π / 2

Third problem:

cot⁴(x) = 4 csc²(x) - 7 Since 1 + cot² x = csc² x, substituting

cot⁴(x) = 4 (1 + cot² x) - 7

cot⁴(x) = 4 + 4 cot² x - 7 Collecting like terms and moving all terms to the left,

cot⁴(x) - 4 cot² x + 3 = 0 Factoring

(cot² x - 3) ( cot² x - 1) = 0

cot² x - 3 = 0 and cot² x - 1 = 0

cot² x = 3 and cot² x = 1 Taking the reciprocals of both sides.

tan² x = 1/3 and tan²x = 1 Taking the square roots and rationalizing.

tan x = ±√3/3 and tan x = ±1 From the unit circle

x = π / 6, 5π / 6, 7π / 6, 11π / 6 and x = π / 4, 3π / 4, 5π / 4, 7π / 4

I am still working on the second problem.

Answer 2

csc² x(csc x + 1) = 4(csc x + 1), you complete

tan (3θ/2) = -3√3, you complete

cot⁴ x = 4(1 + cot²) - 7, combine like terms, use zero-product rule