DOMAIN AND RANGE OF COS INVERSE X

Find all values of x such that

Problem 1 :

-6π ≤ x ≤ 6π and cos x = 0

Solution :

cos x = 0

x = cos^-1(0)

For what angle measure of cosine, we get the value 0.

General solution for cosine function for cos^-1(0) :

Problem 2 :

-5π ≤ x ≤ 5π and cos x = 1

Solution :

cos x = 1

x = cos^-1(1)

For what angle measure of cosine, we get the value 1.

General solution for cosine function for cos^-1(1) :

x = 2nπ, where n ∈ N

If n = 3

x = 2(3)π

x = 6π∉-5π ≤ x ≤ 5π

So, the possible values of n are

0, ±1, ±2, -3

Problem 3 :

State the reason for

Solution :

Here the angle measure (-π/6) mentioned is not in the domain [0, π]. So, it is not true.

Problem 4 :

Is cos^-1(-x) = π - cos^-1(x) true ? justify your answer ?

Solution :

cos^-1(-x) = π - cos^-1(x)

Here the value mention is negative. Using ASTC formula in 2n quadrant and in 3rd quadrant cosine will be negative. Considering the domain [0, π].

To get the angle, we should use

π - cos^-1(x)

So, the given statement is true.

Problem 5 :

Find the principal value of cos^-1 (1/2)

Solution :

cos^-1 (1/2)

For what angle measure of cosine, we get 1/2.

For 60 degree, we get 1/2 and it lies in the domain [0, π].

cos^-1 (1/2) = π/3

Find the value of

Problem 6 :

2cos^-1(1/2) + sin^-1(1/2)

Solution :

= 2cos^-1(1/2) + sin^-1(1/2)

= 2 (π/3) + π/6

= 2π/3 + π/6

= (4π + π)/6

= 5π/6

Problem 7 :

cos^-1(1/2) + sin^-1(-1)

Solution :

For what angle measure of cosine, we get 1/2 and for what angle measure of sin, we get -1.

=  (π/3) +  (3π/2)

Here  (π/3) is in the domain of cosine. But 3π/2 is not in the domain of sin.

So, -π/2 is the replacement of 3π/2.

=  (π/3) +  (-π/2)

= (2π - 3π)/6

= -π/6

Problem 8 :

Solution :

Problem 9 :

Solution :

Combining the above domain, we get [-5, 5]

So, the domain of the function f(x) is [-5, 5].

Problem 10 :

For what value of x, the inequality

π/2 < cos^-1 (3x- 1) < π

holds ?

Solution :

To solve for x,

π/2 < cos^-1 (3x- 1) < π

cos (π/2) < (3x - 1) < cos π

0 < 3x - 1 < -1

Add 1

1 < 3x < 0

Divide by 3

1/3 < x < 0/3

1/3 < x < 0

The inequality is true when,

0 < x < 1/3

Find the value of

Problem 10 :

Solution :

Problem 11 :

Solution :