JEE Advanced Level Test: Trigonometric Equations- 1 - JEE MCQ

# JEE Advanced Level Test: Trigonometric Equations- 1 - JEE MCQ

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## 26 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced Level Test: Trigonometric Equations- 1

JEE Advanced Level Test: Trigonometric Equations- 1 for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The JEE Advanced Level Test: Trigonometric Equations- 1 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Trigonometric Equations- 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Trigonometric Equations- 1 below.
JEE Advanced Level Test: Trigonometric Equations- 1 - Question 1

### The general solution of the equation, 2cos2x = 3.2cos2x – 4 is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 1

2⋅cos2x=2(2cos2x−1) x+cos2x
=2cos2 x − 4
x=2cos2x (cosx=±1)/x
=nπ, n∈I

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 2

### The solution set of the equation 4sinθ . cosθ – 2 cosθ – 2√3 sinθ + √3 = 0 in the interval (0, 2π) is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 2

4sinqcosq - 2cosq - 2√3sinq + √3 = 0
2cosq(2sinq - 1) - √3(2sinq - 1) = 0
(2sinq - 1) (2cosq - √3) = 0
sinq = 1/2 , cosq = √3/2
q = π/6, 5π/6. , q = π/6, 11π/6
q = {π/6 , 5π/6, 11π/6}

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JEE Advanced Level Test: Trigonometric Equations- 1 - Question 3

### Total number of solutions of sin x . tan 4x = cos x belonging to (0, π) are

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 3

sin x tan 4x = cos x

sin x sin 4x = cos x cos 4x

cos x cos 4x - sin x sin 4x = 0

cos(x + 4x) = 0

cos 5x = 0

5x = pi/2 or 5x = 3pi/2 or 5x = 5pi/2

or 5x = 7pi/2 or 5x = 9pi/2

x = pi/10 or x = 3pi/10 or x = pi/2

or x = 7pi/10 or x = 9pi/10.

number of solutions is 5.

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 4

All solutions of the equation, 2 sin θ + tan θ = 0 are obtained by taking all integral values of m and n in

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 4

2 sin θ + tan θ = 0⇒2 sin θ + sin θcos θ = 0⇒2 sin θ . cos θ + sin θ = 0⇒sin θ[2 cos θ + 1] = 0⇒sin θ = 0  or  2 cos θ + 1 = 0⇒sin θ = 0  or cos θ = −12⇒sin θ = 0  or cos θ = −cos (π/3)⇒sin θ = 0  or cos θ = cos (2π/3)⇒θ = nπ or θ = 2mπ±2π3

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 5

The most general solution of tan θ = – 1 and cos θ = 1/√2 is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 5

We have tan θ = -1 and cos θ =1/√2 .
So, θ lies in IV quadrant.
θ = 7/4
So, general solution is θ = 7π/4 + 2 n π, n∈ Z

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 6

If 2 cos2 (π + x) + 3 sin (π + x) vanishes then the values of x lying in the interval from 0 to 2π are

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 6

2cos2(π+x)+3sin(π+x)
=2cos2(x)−3sin(x)
=(1−sin2(x))⋅2−3sin(x)
∴sin(x)=−2,sin(x)=1/2
sin(x)=−2:None
and sin(x) = 1/2: x=π/6+2πn, x=5π/6+2πn
∴x=π/6 or 5π/6

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 7

If x ∈ , the number of solutions of the equation, sin 7x + sin 4x + sin x = 0 is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 7

We have, sin7x+sinx+sin4x=0
2sin4xcos3x+sin4x=0
⇒sin4x(2cos3x+1)=0
⇒sin4x=0,cos3x=−1/2
⇒4x=2kπ
⇒2kπ/4
​=kπ/2
if k=0⇒x=0
if k=1⇒x=π/2
cos3x=−1/2
⇒ Again 3x=±2π/3, 5π/18
3x=2π/3=2kπ
⇒x=2π/9+2kπ/3
k=0, x=2π/9
k=1, x=2π/9+2π/3=8π/9
out of range
x=(−2π)/3 + 2kπ (out of range)
∴ For x=[0,π/2] we get 5
solution:
{0,2π/9,π/2} {2π/3,5π/18}

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 8

The general solution of sin x + sin 5x = sin 2x + sin 4x is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 8

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 9

A triangle ABC is such that sin(2A + B) = 1/2. If A, B, C are in A.P. then the angle A, B, C are respectively

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 9

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 10

if

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 11

if

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 12

If cos 2θ + 3 cos θ = 0 then

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 13

If sin θ + 7 cos θ = 5, then tan (θ/2) is a root of the equation

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 14

sin 3θ = 4 sin θ . sin 2θ . sin 4q in 0 < θ < π has

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 15

General solution of the equation, cot 3 θ – cot θ = 0 is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 15

cot3θ−cotθ=0
(3cotθ−cot3 θ)/(1−3cot2 θ)=cotθ
cotθ(3−cot2 θ)=cotθ(1−3cot2 θ)
⟹ cotθ=0
⟹ cotθ=cotπ/2
⟹ θ=(2n−1)π/2

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 16

The set of values of x for which = 1 is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 16

tan(A-B)=(tanA-tanB)/(1+tanA*tanB) THEN
tan(3x-2x)=( tan3x−tan2x)/(1+tan3xtan2x)
tanx = 1
x =π/4
x =nπ +π/4

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 17

The principal solution set of the equation, 2 cos x =  is

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 18

The number of all possible triplets (a1, a2, a3) such that : a1 + a2 cos 2x + asin2x = 0 for all x is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 18

we can write sin2x=[(1−cos2x)]/2
(a1)+(a3)/2=0
(a2)−(a3)/2=0
These two equations can have infinite roots.

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 19

The value 'a' for which the equation 4cosec2 (π(a + x)) + a2 – 4a = 0 has a real solution is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 19

Dear Sanchit

4 cosec 2 (π (a+x) ) + a2 - 4a = 0

4 cosec 2 (π (a+x) ) =- a2 + 4a

L .H .S is always greater than 4 so for real solution

-a2 + 4a ≥4

or (a-2)2≤0

so only a=2 is possible

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 20

The solution of |cos x| = cosx – 2sinx is

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 20

Squaring both sides

cos^2 x=cos^2 x + 4sin^2x-4sinxcosx

4sinxcox=4sin^2 x

sin x=cos x

Principal solution = pi/4=45 degrees

General Solution = 2nπ+π/4

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 21

The number of solutions of sin θ + 2sin 2θ + 3sin 3θ + 4sin 4θ = 10 in (0, π) is

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 22

The arithmetic mean of the roots of the equation 4cos3x – 4cos2x – cos(π + x) – 1 = 0 in the interval [0, 315] is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 22

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 23

Find the no. of roots of the equation tan x + sec x = 2 cos x in the interval [0, 2π]-

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 24

General solution of tan 5θ = cot 2θ is-

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 25

The number of values of x in the interval [0, 3π]satisfying the equation 2 sin2 x + 5 sin x – 3 = 0 is –

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 25

JEE Advanced Level Test: Trigonometric Equations- 1 - Question 26

If 0 < x < π, and cos x + sin x = 1/2, then tan x is –

Detailed Solution for JEE Advanced Level Test: Trigonometric Equations- 1 - Question 26

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