Trignometric Ratios, Functions and Equations Test

Result

Trignometric Ratios, Functions and Equations Test - 10

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Weekly Quiz Competition

Question 1
2 / -0.83

If sin θ= 1/2. How many solutions does this equation have between 0 and π?

A
2

B
4

C
1

D
3

Solution

Two results are possible as π/6 &5 π/6 comes by subtracting π/6from π, In general we have Sin ∆=n π+(-1)^n(A) here A is π/6 and by putting different values of n in integer 2 values can be obtained
Question 2
2 / -0.83

The general solution of sin = 0 is

A
n πwhere n is a real number

B
n π, where n is an integer

C
2 π

D
π

Solution

sin θ= 0
⇒sin-1 θ= 0,π,2 π,...
⇒θ= 0 + n π→n ∈Z
The general solution for sin x = 0 will be, x = n π, where n ∈I.
Question 3
2 / -0.83

if cos θ= √3/2. How many solutions does this equation have between -πand π?

A
2

B
1

C
3

D
4

Solution

The general solution of the given question is theta= 2n π±π/6 but it is mentioned that they are lies between -πto π. So when we put n=0 we get theta =±π/6. And when we put n= 1 we get theta does not lies between -πto ÷π. So we get only two values of theta.
Question 4
2 / -0.83

The solution of cos 4x = cos 2x

A

B

C

D

Solution

cos 4x = cos 2x
Question 5
2 / -0.83

The solution of sin θcos θ= √3/4 is

A

B

C

D

Solution
Question 6
2 / -0.83

The principal solution for tan x = √3 is,

A
π, 2 π

B
π/3, 4 π/3

C
2 π/3, π/6

D
π/6, 2 π/6

Solution
Question 7
2 / -0.83

The solution of the equation tan 2 θ= tan θ/2 is

A
n π/2 - [(n² π² + 16)^1/2 ]/4

B
n π/2 + [(n² π² + 16)^1/2 ]/4

C
n π/2 +- [(n² π² + 16)^1/2 ]/4

D
n π/2 +- [(n² π² + 16)^1/2 ]/2

Solution

tan θ= tan α
θ= n π+ α
2 θ= n π+ 2/θ
2(θ- 1/θ) = n π
= θ- 1/θ= n π/2
= θ² - (n πθ)/2 - 1 = 0
θ= n π/2 +- [(n² π² + 4)^1/2 ]/2
θ= n π/2 +- [(n² π² + 16)^1/2 ]/4
Question 8
2 / -0.83

If tan x = tan α, then the general solution of the equation is

A
n π–α

B
2n π- α

C
n π+ α

D
2n π+ α

Solution
Question 9
2 / -0.83

tan(π+ x)=

A
-tan x

B
tan π+ tan x

C
0

D
tan x

Solution

(180+theta ) lies in third quadrant...where tantheta is positive....tan(180+x)=tanx
Question 10
2 / -0.83

The most general value that satisfies the equation cosec θ= 2 and cot θ= -√3 is

A
2n π+ 5 π/ 6, n ∈Z

B
2n π±5 π/ 6, n ∈Z

C
2n π+ π/6, n ∈Z

D
2n π–5 π/6, n ∈Z

Solution