# Solving Trigonometric Equations Worksheet

Solving Trigonometric Equations Worksheet
• Page 1
1.
Write the maximum value of 3cos $x$ + 4sin $x$ + 3.
 a. 8 b. 3 c. 5

#### Solution:

The maximum value of acos x + bsin x + c is given as c + a² + b².

Consider, 3cos x + 4sin x + 3

Maximum value = 3 + 3² + 4²
[Use step 1.]

= 3 + 25 = 3 + 5 = 8
[Simplify.]

Therefore, the maximum value of 3cos x + 4sin x + 3 = 8.

2.
Find the general solution of the tan $\theta$ + tan 4$\theta$ + tan $\theta$ tan 4$\theta$ = 1 with $n$ as an integer.
 a. $n$$\pi$ + $\frac{\pi }{5}$ b. $\frac{n\pi }{4}+\frac{\pi }{10}$ c. $\frac{n\pi }{5}+\frac{\pi }{20}$ d. $\frac{n\pi }{5}+\frac{\pi }{4}$

#### Solution:

tan θ + tan 4θ + tan θ tan 4θ = 1

tan θ + tan 4θ = 1 - tan θ tan 4θ
[Subtract tan θ tan 4θ on both the sides.]

Therefore, tan θ+tan 4θ1-tan 4θtan θ = 1
[Divide by 1 - tan θ tan 4θ on both the sides.]

tan θ+tan 4θ1-tan 4θtan θ = tanπ4
[Use tan π4 = 1.]

tan 5θ = tan π4
[Use sum identity tan x+tan y1-tan x tan y = tan(x + y).]

5θ = nπ + π4, where n is an integer.

θ = nπ5+π20
[Divide by 5 on both the sides.]

Therefore, the general solution of tan θ + tan 4θ + tan θ tan 4θ = 1 is θ = nπ5+π20

3.
The value of cos $\frac{\pi }{7}$ cos $\frac{2\pi }{7}$ cos $\frac{4\pi }{7}$ is ______
 a. $\frac{1}{16}$ b. $\frac{1}{8}$ c. $\frac{-1}{8}$ d. $\frac{1}{2}$

#### Solution:

Given cosπ7 cos2π7 cos4π7

2 sinπ7 cosπ7 cos2π7 cos4π72 sinπ7
[Multiply and divide by 2sinπ7.]

sin2π7 cos2π7 cos4π72 sinπ7
[Use 2sinAcosA = sin2A.]

2sin2π7 cos2π7 cos4π74 sinπ7
[Multiply and divide by 2.]

sin4π7 cos4π74 sinπ7
[Use 2sinAcosA = sin2A.]

2sin4π7 cos4π78 sinπ7
[Multiply and divide by 2.]

sin8π78 sinπ7
[Use 2sinAcosA = sin2A.]

sin π+π78 sinπ7
[sin π + θ = - sin θ]

-sinπ78 sinπ7 = -1 / 8
[Simplify.]

4.
Find the general solution of the equation 2(cos $\theta$ + sec $\theta$) = 5.
 a. 2$n$$\pi$ ± $\frac{\pi }{4}$ b. 2$n$$\pi$ ± $\frac{\pi }{3}$ c. 2$n$$\pi$ ± $\frac{\pi }{6}$ d. 2$n$$\pi$ ± $\frac{\pi }{2}$

#### Solution:

2(cos θ + 1cos θ) = 5
[Write 1cos θ for sec θ.]

2(cos² θ + 1) = 5cos θ
[Simplify.]

2cos² θ + 2 = 5 cos θ

2cos² θ - 5 cos θ + 2 = 0
[Express as quadratic equation in cos θ.]

2cos² θ - 4cos θ - cos θ + 2 = 0
[Rewrite - 5cos θ as (- 4cos θ - cos θ).]

2cos θ(cos θ - 2) - (cos θ - 2) = 0
[Factor.]

(cos θ - 2) (2cos θ - 1) = 0

(cos θ - 2) = 0 , (2cos θ - 1) = 0
[Equate each term to zero.]

cos θ = 2, 1 / 2

Cos θ = 1 / 2= cos π3
[As cos θ ≤ 1, cos θ = 2 has no solution.]

The general solution of θ when θ = k is 2nπ ± k.

Therefore, the general solution of the equation when θ = π / 3 is θ = 2nπ ± π3
[Substitute k = π3.]

Hence the general solution of the equation 2(cos θ + sec θ) = 5 is θ = 2nπ ± π3.

5.
If 0 ≤ $x$ ≤ 2$\pi$ then find the value of $x$ for which $y$ = 2 sin $\frac{x}{2}$ has a maximum value.
 a. $\frac{3\pi }{2}$ b. $\pi$ c. 1 d. 2$\pi$

#### Solution:

y = 2 sin x2 has the maximum value 2.
[The maximum value of sin x is 1 and y = 2(1) = 2.]

2 sin x2 = 2
[Substitute y = 2, From step 1.]

sin x2 = 1
[Divide both sides with 2.]

x2 =π2
[If sin θ = 1 then θ = π2 .]

Hence x = π.
[Multiply by 2 on both sides and simplify.]

Therefore the value of x for which y = 2 sin x2 has a maximum value is π.

6.
The solution set of sin 2$x$ < sin $x$ where 0 ≤ $x$ ≤ 2$\pi$ is _______
 a. b. c. d. $\frac{5\pi }{3}$ < x < 2π and

#### Solution:

Given, sin 2x < sin x.

2sin xcos x < sin x
[Use sin 2x = 2sin xcos x.]

2sin xcos x - sin x < 0
[Group the like terms.]

sin x (2cos x - 1) < 0
[Simplify.]

sin x < 0 and 2cos x - 1 > 0
[Case 1.]

sin x < 0 π < x < 2π
[Identify the region.]

2 cos x - 1 > 0 cos x > 1 / 2
[Simplify.]

0 < x < π3 and 5π3 < x < 2π
[Identify the region.]

The common region is 5π3 < x < 2π.
[From steps 6 and 8.]

sin x > 0 and 2cos x - 1 < 0
[Case 2.]

sin x > 0 0 < x < π
[Identify the region.]

2 cos x - 1 < 0 cos x < 1 / 2
[Simplify.]

π3 < x <5π3
[Identify the region.]

The common region is π3 < x < π.
[From steps 11 and 13.]

Therefore, the solution set of sin 2x < sin x is 5π3 < x < 2π and π3 < x < π.
[From steps 9 and 14.]

7.
Write the minimum value of 7cos $x$ + 24sin $x$ + 28.
 a. 5 b. 4 c. 3

#### Solution:

The minimum value of acos x + bsin x + c is given as c - a² + b².

7cos x + 24sin x + 28

Minimum value = 28 - 7² + 24²
[Use step 1.]

= 28 - 625 = 28 - 25 = 3
[Simplify.]

Therefore, the minimum value of 7cos x + 24sin x + 28 = 3.

8.
The value of the expression $\sqrt{3}$ cosec 20° - sec 20° is equal to _______
 a. 4 b. 2 c. d.

#### Solution:

Given 3 cosec 20° - sec 20°

= 3sin 20° -1cos 20°
[Use cosec θ = 1sin θ and sec θ = 1cos θ.]

= 3 cos 20° - sin 20°sin 20° cos 20°
[Simplify.]

= 2[32 cos 20° -12sin 20°]22 sin 20° cos 20°
[Multiply and divide the numerator and denominator by 2.]

= 4[32 cos 20° -12sin 20°]sin 40°
[Use 2sin A cos A = sin 2A.]

= 4(sin 60° cos 20° - cos 60° sin 20°)sin 40°
[Use 32 as sin 60° and 1 / 2as cos 60°.]

4 [sin (60° - 20°)]sin 40°
[Use sin A cos B - cosA sin B = sin (A - B).]

4 (sin 40°)sin 40° = 4
[Simplify.]

Therefore the value of the expression 3 cosec 20° - sec 20° is 4.

9.
If (1 + tan A)(1 + tan B) = 2, then find (A + B).
 a. $\frac{\pi }{6}$ b. $\frac{\pi }{2}$ c. $\frac{\pi }{3}$ d. $\frac{\pi }{4}$

#### Solution:

(1 + tan A)(1 + tan B) = 2
[Given.]

1 + tan A + tan B + tan A tan B = 2
[Expand.]

tan A + tan B + tan A tan B = 2 - 1 = 1
[Subtract 1 from both the sides.]

tan A + tan B = 1 - tan A tan B
[Subtract tan A tan B from both the sides.]

tan A + tan B1 - tan A tan B = 1
[Divide both the sides by 1 - tan A tan B.]

tan (A + B) = 1
[Use Sum Identity.]

A + B = π4
[tan π4 = 1.]

Therefore, the value of (A + B) = π4.

10.
Write the number of solutions of the equation sin 7$x$ - cosec 7$x$ = 0, if 0° ≤ $x$ ≤ 360°.
 a. 3 b. 2 c. 1

#### Solution:

sin 7x - cosec 7x = 0

sin 7x = cosec 7x

sin27x = 1
[Use cosec x = 1sin x.]

So, sin 7x = +1, -1

x = (90 / 7) °, (270 / 7) °
[Solve for x, such that [0° ≤ x ≤ 360°].]

So, the number of solutions = 2.