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Trigonometric Equations

JEE Mathematics · 41 questions · Page 1 of 5 · Click an option or "Show Solution" to reveal answer

Q1
If $$\alpha,-\frac{\pi}{2}
A 10−1012\dfrac{10-\sqrt{10}}{12}
B 10−106\dfrac{\sqrt{10}-10}{6}
C 10−1012\dfrac{\sqrt{10}-10}{12}
D 10−106\dfrac{10-\sqrt{10}}{6}
Correct Answer
Option C
Solution
4+5tan⁡θ=sec⁡θ4+5 \tan \theta=\sec \theta

Squaring :

24tan⁡2θ+40tan⁡θ+15=024 \tan ^2 \theta+40 \tan \theta+15=0
tan⁡θ=−10±1012\tan \theta=\frac{-10 \pm \sqrt{10}}{12}

and

tan⁡θ=−(10+1012)\tan \theta=-\left(\frac{10+\sqrt{10}}{12}\right)

is Rejected. (3) is correct.

Q2
The number of solutions of the equation 1 + sin4 x = cos23x, x∈[−5π2,5π2]x \in \left[ { - {{5\pi } \over 2},{{5\pi } \over 2}} \right] is :
A 5
B 3
C 7
D 4
Correct Answer
Option A
Solution
1+sin⁡4x‾≥1=cos⁡23x‾≤1\mathop {\underline {1 + {{\sin }^4}x} }\limits_{ \ge 1} = \mathop {\underline {{{\cos }^2}3x} }\limits_{ \le 1}

Now for equality to hold sin4x = 0 & cos23x = 1 sin4 = 0 ⇒\Rightarrow x = -2π\pi, - π\pi, 0, π\pi, 2π\pi All of which satisfy cos23x = 1 ⇒\Rightarrow 5 solutions

Q3
The number of elements in the set S={x∈R:2cos⁡(x2+x6)=4x+4−x}S=\left\{x \in \mathbb{R}: 2 \cos \left(\frac{x^{2}+x}{6}\right)=4^{x}+4^{-x}\right\} is :
A 1
B 3
C 0
D infinite
Correct Answer
Option A
Solution

Given,

2cos⁡(x2+x6)=4x+4−x2\cos \left( {{{{x^2} + x} \over 6}} \right) = {4^x} + {4^{ - x}}

We know, A.M ≥\ge G.M ∴\therefore for 4x and 4−-x

4x+4−x2≥4x . 4−x{{{4^x} + {4^{ - x}}} \over 2} \ge \sqrt {{4^x}\,.\,{4^{ - x}}}
⇒4x+4−x2≥1\Rightarrow {{{4^x} + {4^{ - x}}} \over 2} \ge 1
⇒4x+4−x≥2\Rightarrow {4^x} + {4^{ - x}} \ge 2

...... (1) And we know,

−1≤cos⁡x≤1- 1 \le \cos x \le 1

∴\therefore

−1≤cos⁡(x2+x6)≤1- 1 \le \cos \left( {{{{x^2} + x} \over 6}} \right) \le 1
−2≤2cos⁡(x2+x6)≤2- 2 \le 2\cos \left( {{{{x^2} + x} \over 6}} \right) \le 2

...... (2) (1) and (2) both satisfies only when both equal to 2. ∴\therefore

2cos⁡(x2+x6)=22\cos \left( {{{{x^2} + x} \over 6}} \right) = 2
⇒cos⁡(x2+x6)=1\Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = 1
⇒cos⁡(x2+x6)=cos⁡0\Rightarrow \cos \left( {{{{x^2} + x} \over 6}} \right) = \cos 0
⇒x2+x6=0\Rightarrow {{{x^2} + x} \over 6} = 0
⇒x2+x=0\Rightarrow {x^2} + x = 0
⇒x(x+1)=0\Rightarrow x(x + 1) = 0
⇒x=0,−1\Rightarrow x = 0, - 1

When

x=0x = 0

, L.H.S

=2cos⁡(x2+x6)= 2\cos \left( {{{{x^2} + x} \over 6}} \right)
=2cos⁡(06)= 2\cos \left( {{0 \over 6}} \right)
=2cos⁡0= 2\cos 0
=2 . 1= 2\,.\,1
=2= 2

R.H.S

=4x+4−x= {4^x} + {4^{ - x}}
=40+40= {4^0} + {4^0}
=2= 2

∴\therefore

x=0x = 0

is accepted. Now, when

x=−1x = - 1

, L.H.S

=2cos⁡(x2+x6)= 2\cos \left( {{{{x^2} + x} \over 6}} \right)
=2cos⁡(1−16)= 2\cos \left( {{{1 - 1} \over 6}} \right)
=2cos⁡0=2= 2\cos 0 = 2

R.H.S

=4x+4−x= {4^x} + {4^{ - x}}
=4−1+41= {4^{ - 1}} + {4^1}
=14+4= {1 \over 4} + 4
=174= {{17} \over 4}

∴\therefore L.H.S ≠\ne R.H.S ∴\therefore

x=−1x = - 1

is not a solution. ∴\therefore Only one solution possible which is

x=0x = 0

.

Q4
The sum of all values of θ\theta ∈ \in (0,π2)\left( {0,{\pi \over 2}} \right) satisfying sin2 2θ\theta + cos4 2θ\theta = 34{3 \over 4} is -
A 5π4{{5\pi } \over 4}
B π2{\pi \over 2}
C π\pi
D 3π8{{3\pi } \over 8}
Correct Answer
Option B
Solution

sin22θ\theta + cos42θ\theta =

34,{3 \over 4},

θ\theta

∈\in
(0,π2)\left( {0,{\pi \over 2}} \right)

⇒\Rightarrow 1 −- cos22θ\theta + cos42θ\theta =

34{3 \over 4}

⇒\Rightarrow 4cos2θ\theta −- 4cos22θ\theta + 1 = 0 ⇒\Rightarrow (2cos22θ\theta −- 1)2 = 0 ⇒\Rightarrow cos22θ\theta =

12{1 \over 2}

= cos2

π4{{\pi \over 4}}

⇒\Rightarrow 2θ\theta = nπ\pi ±\pm

π4{\pi \over 4}

, n

∈\in

I ⇒\Rightarrow θ\theta =

nπ2±π8{{n\pi } \over 2} \pm {\pi \over 8}

⇒\Rightarrow θ\theta =

π8,π2−π8{\pi \over 8},{\pi \over 2} - {\pi \over 8}

Sum of solutions

π2{\pi \over 2}
Q5
Let S = {θ\theta ∈ \in [–2π\pi , 2π\pi ] : 2cos2θ\theta + 3sinθ\theta = 0}. Then the sum of the elements of S is
A π\pi
B 2π\pi
C 13π6{{13\pi } \over 6}
D 5π3{{5\pi } \over 3}
Correct Answer
Option B
Solution

2cos2θ\theta + 3sinθ\theta = 0 ⇒\Rightarrow 2(1 - sin2θ\theta) + 3sinθ\theta = 0 ⇒\Rightarrow 2sin2θ\theta - 3sinθ\theta - 2 = 0 ⇒\Rightarrow 2sin2θ\theta - 4sinθ\theta + sinθ\theta - 2 = 0 ⇒\Rightarrow (2sinθ\theta + 1)(sinθ\theta - 2) = 0 ∴\therefore sinθ\theta = 2 (Not possible) sinθ\theta =

−12- {1 \over 2}

∴\therefore θ\theta = nπ\pi + (-1)n

(−π6)\left( { - {\pi \over 6}} \right)

When n = 0, θ\theta =

−π6{ - {\pi \over 6}}

When n = 1, θ\theta =

7π6{{7\pi } \over 6}

When n = -1, θ\theta =

−5π6{-{5\pi } \over 6}

When n = 2, θ\theta =

11π6{{11\pi } \over 6}

∴\therefore Sum of all solutions in [–2π\pi, 2π\pi] is =

−π6+7π6−5π6+11π6- {\pi \over 6} + {{7\pi } \over 6} - {{5\pi } \over 6} + {{11\pi } \over 6}

= 2π\pi

Q6
If [x] denotes the greatest integer ≤ \le x, then the system of linear equations [sin θ\theta ]x + [–cosθ\theta ]y = 0, [cotθ\theta ]x + y = 0
A has a unique solution if θ∈(π2,2π3)\theta \in \left( {{\pi \over 2},{{2\pi } \over 3}} \right) and have infinitely many solutions if θ∈(π,7π6)\theta \in \left( {\pi ,{{7\pi } \over 6}} \right)
B have infinitely many solutions if θ∈(π2,2π3)\theta \in \left( {{\pi \over 2},{{2\pi } \over 3}} \right) and has a unique solution if θ∈(π,7π6)\theta \in \left( {\pi ,{{7\pi } \over 6}} \right)
C have infinitely many solutions if θ∈(π2,2π3)∪(π,7π6)\theta \in \left( {{\pi \over 2},{{2\pi } \over 3}} \right) \cup \left( {\pi ,{{7\pi } \over 6}} \right)
D has a unique solution if θ∈(π2,2π3)∪(π,7π6)\theta \in \left( {{\pi \over 2},{{2\pi } \over 3}} \right) \cup \left( {\pi ,{{7\pi } \over 6}} \right)
Correct Answer
Option B
Solution

if θ\theta

∈\in
(π2,2π3)\left( {{\pi \over 2},{{2\pi } \over 3}} \right)
⇒cos⁡θ∈(−12,0)⇒−cos⁡θ∈(0,12)\Rightarrow \cos \theta \in \left( { - {1 \over 2},0} \right) \Rightarrow - \cos \theta \in \left( {0,{1 \over 2}} \right)
sin⁡θ∈(32,1)\sin \theta \in \left( {{{\sqrt 3 } \over 2},1} \right)

and

cot⁡θ∈(−13,0)\cot \theta \in \left( { - {1 \over {\sqrt 3 }},0} \right)

If

θ∈(π,7π6)⇒cos⁡θ∈(−1,−32)⇒−cos⁡θ∈(32,1)\theta \in \left( {\pi ,{{7\pi } \over 6}} \right) \Rightarrow \cos \theta \in \left( { - 1,{{ - \sqrt 3 } \over 2}} \right) \Rightarrow - \cos \theta \in \left( {{{\sqrt 3 } \over 2},1} \right)
sin⁡θ∈(−12,0)\sin \theta \in \left( { - {1 \over 2},0} \right)

and

cot⁡θ∈(3,∞)\cot \theta \in \left( {\sqrt 3 ,\infty } \right)

Then in θ\theta

∈\in
(π2,2π3)\left( {{\pi \over 2},{{2\pi } \over 3}} \right)

⇒\Rightarrow [sinθ\theta] = 0; [-cosθ\theta] = 0; [cotθ\theta] = -1; Hence 0 = 0 and -x + y = 0 [have infinitely solutions] and in

θ∈(π,7π6)\theta \in \left( {\pi ,{{7\pi } \over 6}} \right)

⇒\Rightarrow [sinθ\theta] = -1; [-cosθ\theta] = 0; [cotθ\theta] = 1, 2.

3 ...........; Then -x - 0.y = 0 ⇒\Rightarrow x = 0

[cot⁡θ]x+y=0[\cot \theta ]x + y = 0

⇒\Rightarrow 1.x + y = 0 or 2x + y = 0 or ....... each of the line will cut x = 0 at exactly one point.

Hence unique solutions.

Q7
The number of solution of tan⁡ x+sec⁡ x=2cos⁡ x\tan \,x + \sec \,x = 2\cos \,x in [0, 2 π]\left[ {0,\,2\,\pi } \right] is
A 2
B 3
C 0
D 1
Correct Answer
Option B
Solution

We can simplify the equation by converting everything to sines and cosines:

sin⁡xcos⁡x+1cos⁡x=2cos⁡x\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x

Multiplying through by cos⁡x\cos x gives:

sin⁡x+1=2cos⁡2x\sin x + 1 = 2\cos^2 x

Using the identity cos⁡2x+sin⁡2x=1\cos^2 x + \sin^2 x = 1, we can substitute sin⁡2x\sin^2 x with 1−cos⁡2x1 - \cos^2 x to get:

sin⁡x+1=2−2sin⁡2x\sin x + 1 = 2 - 2\sin^2 x

Rearranging terms gives:

2sin⁡2x+sin⁡x−1=02\sin^2 x + \sin x - 1 = 0

This is a quadratic equation in sin⁡x\sin x, which we can solve using the quadratic formula:

sin⁡x=−1±1+84\sin x = \frac{-1 \pm \sqrt{1 + 8}}{4}

The discriminant is positive, so there are two solutions for sin⁡x\sin x:

sin⁡x=−1±94=−1,12\sin x = \frac{-1 \pm \sqrt{9}}{4} = -1, \frac{1}{2}

For sin⁡x=−1\sin x = -1, we have x=3π2x = \dfrac{3\pi}{2}.

For sin⁡x=12\sin x = \dfrac{1}{2}, we have x=π6,5π6x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}.

Therefore, there are a total of 3\boxed{3} solutions in the interval [0,2π]\left[ 0, 2\pi \right].

Q8
The number of values of xx in the interval [0,3π] \left[ {0,3\pi } \right]\, satisfying the equation 2sin⁡2x+5sin⁡x−3=02{\sin ^2}x + 5\sin x - 3 = 0 is
A 4
B 6
C 1
D 2
Correct Answer
Option A
Solution
2sin⁡2x+5sin⁡x−3=02{\sin ^2}x + 5\sin x - 3 = 0
⇒(sin⁡x+3)(2sin⁡x−1)=0\Rightarrow \left( {\sin x + 3} \right)\left( {2\sin x - 1} \right) = 0
sin⁡x=12\sin x = {1 \over 2}

and

  sin⁡x≠−3\,\,\sin x \ne - 3

Given that

x∈[0,3π]x \in \left[ {0,3\pi } \right]

So possible values of x are

30∘30^\circ

,

150∘150^\circ

,

390∘390^\circ

,

510∘510^\circ

. That means x have 4 values.

Q9
If 0≤x<2π0 \le x < 2\pi , then the number of real values of xx, which satisfy the equation  cos⁡x+cos⁡2x+cos⁡3x+cos⁡4x=0\,\cos x + \cos 2x + \cos 3x + \cos 4x = 0 is:
A 7
B 9
C 3
D 5
Correct Answer
Option A
Solution
cos⁡x+cos⁡2x+cos⁡3x+cos⁡4x=0\cos x + \cos 2x + \cos 3x + \cos 4x = 0

⇒\Rightarrow

(cos⁡x+cos⁡3x)(\cos x + \cos 3x)

+

(cos⁡2x+cos⁡4x)(\cos 2x + \cos 4x)

= 0

⇒2cos⁡2xcos⁡x+2cos⁡3xcos⁡x=0\Rightarrow 2\cos 2x\cos x + 2\cos 3x\cos x = 0
⇒2cos⁡x(2cos⁡5x2cos⁡x2)=0\Rightarrow 2\cos x\left( {2\cos {{5x} \over 2}\cos {x \over 2}} \right) = 0
cos⁡x=0,cos⁡5x2=0,cos⁡x2=0\cos x = 0,\cos {{5x} \over 2} = 0,\cos {x \over 2} = 0
x=π,π2,3π2,π5,3π5,7π5,9π5x = \pi ,{\pi \over 2},{{3\pi } \over 2},{\pi \over 5},{{3\pi } \over 5},{{7\pi } \over 5},{{9\pi } \over 5}
Q10
If sum of all the solutions of the equation 8cos⁡x.(cos⁡(π6+x).cos⁡(π6−x)−12)=18\cos x.\left( {\cos \left( {{\pi \over 6} + x} \right).\cos \left( {{\pi \over 6} - x} \right) - {1 \over 2}} \right) = 1 in [0, π\pi ] is kπ\pi , then k is equal to
A 209{{20} \over 9}
B 23{2 \over 3}
C 139{{13} \over 9}
D 89{{8} \over 9}
Correct Answer
Option C
Solution

As we know,

cos⁡(x+y)cos⁡(x−y)=cos⁡2x−sin⁡2y\cos \left( {x + y} \right)\cos \left( {x - y} \right) = {\cos ^2}x - {\sin ^2}y

∴\therefore

cos(π6+x)cos(π6−x)=cos⁡2(π6)−sin⁡2x{\mathop{\rm cos}\nolimits} \left( {{\pi \over 6} + x} \right)cos\left( {{\pi \over 6} - x} \right) = {\cos ^2}\left( {{\pi \over 6}} \right) - {\sin ^2}x

Given,

8cos⁡x(cos⁡(π6+x)cos⁡(π6−x)−12)=18\cos x\left( {\cos \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right) - {1 \over 2}} \right) = 1
⇒8cos⁡x(cos⁡2π6−sin⁡2x−12)=1\Rightarrow 8\cos x\left( {{{\cos }^2}{\pi \over 6} - {{\sin }^2}x - {1 \over 2}} \right) = 1
⇒8cos⁡x(34−12−sin⁡2x)=1\Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - {{\sin }^2}x} \right) = 1
⇒8cos⁡x(34−12−1+cos⁡2x)=1\Rightarrow 8\cos x\left( {{3 \over 4} - {1 \over 2} - 1 + {{\cos }^2}x} \right) = 1
⇒8cos⁡x(cos⁡2x−34)=1\Rightarrow 8\cos x\left( {{{\cos }^2}x - {3 \over 4}} \right) = 1
⇒2(4cos⁡3x−3cos⁡x)=1\Rightarrow 2\left( {4{{\cos }^3}x - 3\cos x} \right) = 1

(Taking

4cos⁡x4\cos x

inside the bracket). We know,

4cos⁡3x−3cos⁡x=cos⁡3x4{\cos ^3}x - 3\cos x = \cos 3x
⇒2cos⁡3x=1\Rightarrow 2\cos 3x = 1
⇒cos⁡3x=12\Rightarrow \cos 3x = {1 \over 2}

∴\therefore

   \,\,\,
3x=2nπ±π33x = 2n\pi \pm {\pi \over 3}

So,

x=2nπ3±π9x = {{2n\pi } \over 3} \pm {\pi \over 9}

At

n=0,n=0,
x=+π9x = + {\pi \over 9}

as

x∈[0,π]x \in \left[ {0,\pi } \right]

At

n=1,n=1,
x=2π3+π9x = {{2\pi } \over 3} + {\pi \over 9}

∴\therefore

   \,\,\,
x=2π3+π9x = {{2\pi } \over 3} + {\pi \over 9}

and

x=2π3−π9x = {{2\pi } \over 3} - {\pi \over 9}

At

n=2,n = 2,
x=4π3±π9x = {{4\pi } \over 3} \pm {\pi \over 9}

Which does not belongs to

[0,π]\left[ {0,\pi } \right]

So, possible values of

xx

is

=π9,2π3+π9,2π3−π3= {\pi \over 9},{{2\pi } \over 3} + {\pi \over 9},{{2\pi } \over 3} - {\pi \over 3}

Sum of the solutions

=π9+2π3+π9+2π3−π9= {\pi \over 9} + {{2\pi } \over 3} + {\pi \over 9} + {{2\pi } \over 3} - {\pi \over 9}

v

=4π3+π9= {{4\pi } \over 3} + {\pi \over 9}
=13 π9= {{13\,\pi } \over 9}

According to the question,

kπ=13π9k\pi = {{13\pi } \over 9}
∴   \therefore\,\,\,
k=139k = {{13} \over 9}
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