help trigonometric-equations.

Question

TRIG EQUATIONS

 

1) Solve each equation for 0o < x < 3600 (round to nearest tenth of degree).

a) sin x = -1/3

b) cot x = -√3/3

 

2) Solve each equation for 0 ≤ x < 2π (round to the nearest hundredth of a radian).

a) sin x =√2/2

b) cos x = -2/3

 

3) Simplify.

a)sec x/tan x

b) (sec x – 1)(sec x + 1)

c) (sin x)(sec x)

 

Answer 1

(1)

(a) The expression is sin(x) = -1/3

The general solution of sine function is .

The solution of sin(x) = -1/3 is

Substitute different values of n.

If n = 0 then x = -19.47 which is not in the range of 0°< x < 360°.

If n = 1 then x = 199.47 which is in the range of 0°< x < 360°.

If n = 2 then x = 340.53 which is in the range of 0°< x < 360°.

If n = 3 then x = 559.47 which is not in the range of 0°< x < 360°.

Therefore the solutiuons of x are 199.47° and 340.53°.

Answer 2

(1)

(b) The expression is cot(x) = -√3/3

Re-write the expression.

cot(x) = -1/√3

The general solution of cotangent function is .

The solution of cot(x) = -1/√3 is

Substitute different values of n.

If n = 0 then x = 120 which is in the range of 0°< x < 360°.

If n = 1 then x = 300 which is in the range of 0°< x < 360°.

If n = 2 then x = 480 which is not in the range of 0°< x < 360°.

Therefore the solutiuons of x are 120° and 300°.

Answer 3

(2)

(a) The expression is sin(x) = -√2/2

Re-write the expression.

sin(x) = -1/√2

The general solution of sine function is .

The solution of sin(x) = -1/√2 is

Convert the above soltuion into radians.

Substitute different values of n.

If n = 0 then x = which is not in the range of 0 < x < 2π.

If n = 1 then x = which is in the range of 0 < x < 2π.

If n = 2 then x = which is in the range of 0 < x < 2π.

If n = 3 then x = which is not in the range of 0 < x < 2π.

Therefore the solutiuons of x are and .

Answer 4

(2)

(b) The expression is cos(x) = -2/3

The general solution of cosine function is .

The solution of cos(x) = -2/3 is

Convert the function into radians.

x = 2nπ ± 0.7322π

Substitute different values of n.

If n = 0 then

x = -0.7322π which is not in the range of 0 < x < 2π.

x = 0.7322π which is in the range of 0 < x < 2π.

 

If n = 1 then

x = 1.2678π which is in the range of 0 < x < 2π.

x = 2.7322π which is not in the range of 0 < x < 2π.

 

If n = 2 then

x =3.2678 which is not in the range of 0 < x < 2π.

Therefore the solutiuons of x are 0.73π and 1.27π.

Answer 5

3a)

The trigonometric function is sec x / tan x .

Rewrite sec x = ( 1 / cos x )  and tan x  = ( sin x / cos x ) .

Cancel common terms .

Answer 6

3b)

The trigonometric function is (sec x – 1)(sec x + 1) .

The above expression is in the form of  (a+b)(a-b) = a² - b² .

(sec x – 1)(sec x + 1) = sec² x  - 1² .

(sec x – 1)(sec x + 1) = sec² x  - 1 .

We know the trigonometric identity sec² x  - 1 = tan² x .

Therefore (sec x – 1)(sec x + 1) = tan² x .

Answer 7

3c)

The trigonometric function is (sin x)(sec x) .

Rewrite sec x = ( 1 / cos x ) .

We know that  ( sin x / cos x ) = tan x  .