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Trigonometric Equations

JEE Mathematics · 41 questions · Page 2 of 5 · Click an option or "Show Solution" to reveal answer

Q11
The number of solutions of sin3x = cos 2x, in the interval (π2,π)\left( {{\pi \over 2},\pi } \right) is :
A 1
B 2
C 3
D 4
Correct Answer
Option A
Solution

sin 3x = cos 2x ⇒\Rightarrow

   \,\,\,

3 sin x −- 4 sin3 x = 1 −- 2 sin2 x ⇒\Rightarrow

   \,\,\,

4 sin3 x −- 2 sin2 x −- 3 sin x + 1 = 0 ⇒\Rightarrow

   \,\,\,

sin x = 1,

−2±258{{ - 2 \pm 2\sqrt 5 } \over 8}

In the interval

(π2,π)\left( {{\pi \over 2},\pi } \right)

, sin x =

−2+258{{ - 2 + 2\sqrt 5 } \over 8}

So, there is only one solution.

Q12
Let S be the set of all α\alpha ∈ \in R such that the equation, cos2x + α\alpha sinx = 2α\alpha – 7 has a solution. Then S is equal to :
A [2, 6]
B [3, 7]
C [1, 4]
D R
Correct Answer
Option A
Solution

1 - 2sin2x + α\alpha sin x = 2α\alpha - 7 ⇒\Rightarrow 2sin2x - α\alpha sin x + (2α\alpha - 8) = 0

sin⁡x=α±α2−8(2α−8)4\sin x = {{\alpha \pm \sqrt {{\alpha ^2} - 8(2\alpha - 8)} } \over 4}
⇒α±α2−16α+644=α±(α−8)4\Rightarrow {{\alpha \pm \sqrt {{\alpha ^2} - 16\alpha + 64} } \over 4} = {{\alpha \pm (\alpha - 8)} \over 4}
α±(α−8)4⇒2α−84,2{{\alpha \pm (\alpha - 8)} \over 4} \Rightarrow {{2\alpha - 8} \over 4},2
⇒α−42,2\Rightarrow {{\alpha - 4} \over 2},2

(rejected) To exists solutions

−1≤α−42≤1⇒−2≤α−4≤2⇒2≤α≤6- 1 \le {{\alpha - 4} \over 2} \le 1 \Rightarrow - 2 \le \alpha - 4 \le 2 \Rightarrow 2 \le \alpha \le 6

∴\therefore

α∈[2,6]\alpha \in \left[ {2,6} \right]
Q13
Let for some real numbers α\alpha and β\beta, a=α−iβa = \alpha - i\beta . If the system of equations 4ix+(1+i)y=04ix + (1 + i)y = 0 and 8(cos⁡2π3+isin⁡2π3)x+a‾y=08\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)x + \overline a y = 0 has more than one solution, then αβ{\alpha \over \beta } is equal to
A −23 - 2\sqrt 3
B 2−32 - \sqrt 3
C 2+32 + \sqrt 3
D −2−3 - 2 - \sqrt 3
Correct Answer
Option B
Solution

Given

a=α−iβa = \alpha - i\beta

and

4ix+(1+i)y=04ix + (1 + i)y = 0

...... (i)

8(cos⁡2π3+isin⁡2π3)x+a‾y=08\left( {\cos {{2\pi } \over 3} + i\sin {{2\pi } \over 3}} \right)x + \overline a y = 0

.... (ii) By (i)

xy=−(1+i)4i{x \over y} = {{ - (1 + i)} \over {4i}}

...... (iii) By (ii)

xy=−a‾8(−12+3i2){x \over y} = {{ - \overline a } \over {8\left( {{{ - 1} \over 2} + {{\sqrt 3 i} \over 2}} \right)}}

..... (iv) Now by (iii) and (iv)

1+i4i=a‾4(−1+3i){{1 + i} \over {4i}} = {{\overline a } \over {4\left( { - 1 + \sqrt 3 i} \right)}}
⇒a‾=(3−1)+(3+1)i\Rightarrow \overline a = \left( {\sqrt 3 - 1} \right) + \left( {\sqrt 3 + 1} \right)i
⇒α+iβ=(3−1)+(3+1)i\Rightarrow \alpha + i\beta = \left( {\sqrt 3 - 1} \right) + \left( {\sqrt 3 + 1} \right)i

∴\therefore

αβ=3−13+1=2−3{\alpha \over \beta } = {{\sqrt 3 - 1} \over {\sqrt 3 + 1}} = 2 - \sqrt 3
Q14
All possible values of θ\theta ∈\in [0, 2π\pi] for which sin 2θ\theta + tan 2θ\theta > 0 lie in :
A (0,π4)∪(π2,3π4)∪(3π2,11π6)\left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {{{3\pi } \over 2},{{11\pi } \over 6}} \right)
B (0,π2)∪(π,3π2)\left( {0,{\pi \over 2}} \right) \cup \left( {\pi ,{{3\pi } \over 2}} \right)
C (0,π2)∪(π2,3π4)∪(π,7π6)\left( {0,{\pi \over 2}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {\pi ,{{7\pi } \over 6}} \right)
D (0,π4)∪(π2,3π4)∪(π,5π4)∪(3π2,7π4)\left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {\pi ,{{5\pi } \over 4}} \right) \cup \left( {{{3\pi } \over 2},{{7\pi } \over 4}} \right)
Correct Answer
Option D
Solution
sin⁡2θ+tan⁡2θ>0\sin 2\theta + \tan 2\theta > 0
⇒sin⁡2θ+sin⁡2θcos⁡2θ>0\Rightarrow \sin 2\theta + {{\sin 2\theta } \over {\cos 2\theta }} > 0
⇒sin⁡2θ(cos⁡2θ+1)cos⁡2θ>0⇒tan⁡2θ(2cos⁡2θ)>0\Rightarrow \sin 2\theta {{(\cos 2\theta + 1)} \over {\cos 2\theta }} > 0 \Rightarrow \tan 2\theta (2{\cos ^2}\theta ) > 0

Note :

cos⁡2θ≠0\cos 2\theta \ne 0
⇒1−2sin⁡2θ≠θ⇒sin⁡θ≠±12\Rightarrow 1 - 2{\sin ^2}\theta \ne \theta \Rightarrow \sin \theta \ne \pm {1 \over {\sqrt 2 }}

Now,

tan⁡2θ(1+cos⁡2θ)>0\tan 2\theta (1 + \cos 2\theta ) > 0
⇒tan⁡2θ>0\Rightarrow \tan 2\theta > 0

(as

cos⁡2θ+1>0\cos 2\theta + 1 > 0

)

⇒2θ∈(0,π2)∪(π,3π2)∪(2π,5π2)∪(3π,7π2)\Rightarrow 2\theta \in \left( {0,{\pi \over 2}} \right) \cup \left( {\pi ,{{3\pi } \over 2}} \right) \cup \left( {2\pi ,{{5\pi } \over 2}} \right) \cup \left( {3\pi ,{{7\pi } \over 2}} \right)
⇒θ∈(0,π4)∪(π2,3π4)∪(π,5π4)∪(3π2,7π4)\Rightarrow \theta \in \left( {0,{\pi \over 4}} \right) \cup \left( {{\pi \over 2},{{3\pi } \over 4}} \right) \cup \left( {\pi ,{{5\pi } \over 4}} \right) \cup \left( {{{3\pi } \over 2},{{7\pi } \over 4}} \right)
Q15
The number of roots of the equation, (81)sin2x + (81)cos2x = 30 in the interval [ 0, π\pi ] is equal to :
A 2
B 3
C 4
D 8
Correct Answer
Option C
Solution
(81)sin⁡2x+(81)1−sin⁡2x=30{(81)^{{{\sin }^2}x}} + {(81)^{1 - {{\sin }^2}x}} = 30
(81)sin⁡2x+81(81)sin⁡2x=30{(81)^{{{\sin }^2}x}} + {{81} \over {{{(81)}^{{{\sin }^2}x}}}} = 30

Let

(81)sin⁡2x=t{(81)^{{{\sin }^2}x}} = t
t+81t=30⇒t2+81=30tt + {{81} \over t} = 30 \Rightarrow {t^2} + 81 = 30t
t2−30t+81=0{t^2} - 30t + 81 = 0
t2−27t−3t+81=0{t^2} - 27t - 3t + 81 = 0
(t−3)(t−27)=0(t - 3)(t - 27) = 0
t=3,27t = 3,27
(81)sin⁡2x=3,33{(81)^{{{\sin }^2}x}} = 3,{3^3}
34sin⁡2x=31,33{3^{4{{\sin }^2}x}} = {3^1},{3^3}
4sin⁡2x=1,34{\sin ^2}x = 1,3
sin⁡2x=14,34{\sin ^2}x = {1 \over 4},{3 \over 4}

in [0, π\pi ] sin x > 0

sin⁡x=12,32\sin x = {1 \over 2},{{\sqrt 3 } \over 2}
x=π6,5π6,π3,2π3x = {\pi \over 6},{{5\pi } \over 6},{\pi \over 3},{{2\pi } \over 3}

Number of solution = 4

Q16
The number of solutions of the equation x + 2tanx = π2{\pi \over 2} in the interval [0, 2π\pi] is :
A 4
B 3
C 2
D 5
Correct Answer
Option B
Solution
x+2tan⁡x=π2x + 2\tan x = {\pi \over 2}

in [0, 2π\pi]

2tan⁡x=π2−x2\tan x = {\pi \over 2} - x
2tan⁡x=π2−x2\tan x = {\pi \over 2} - x
tan⁡x=π4−x2\tan x = {\pi \over 4} - {x \over 2}
y=tan⁡xy = \tan x

and

y=−x2+π4y = {{ - x} \over 2} + {\pi \over 4}

3 intersection points on the graph. ∴\therefore 3 solutions.

Q17
The number of solutions of sin7x + cos7x = 1, x∈\in [0, 4π\pi] is equal to
A 11
B 7
C 5
D 9
Correct Answer
Option C
Solution

sin7x ≤\le sin2x ≤\le 1 ...... (1) and cos7x ≤\le cos2x ≤\le 1 ..... (2) also sin2x + cos2x = 1 ⇒\Rightarrow equality must hold for (1) & (2) ⇒\Rightarrow sin7x = sin2x & cos7x = cos2x ⇒\Rightarrow sin x = 0 & cos x = 1 or cos x = 0 & sin x = 1 ⇒\Rightarrow x = 0, 2π\pi, 4π\pi,

π2,5π2{\pi \over 2},{{5\pi } \over 2}

⇒\Rightarrow 5 solutions

Q18
The sum of all values of x in [0, 2π\pi], for which sin x + sin 2x + sin 3x + sin 4x = 0, is equal to :
A 8π\pi
B 11π\pi
C 12π\pi
D 9π\pi
Correct Answer
Option D
Solution
(sin⁡x+sin⁡4x)+(sin⁡2x+sin⁡3x)=0(\sin x + \sin 4x) + (\sin 2x + \sin 3x) = 0
⇒2sin⁡5x2{cos⁡3x2+cos⁡x2}=0\Rightarrow 2\sin {{5x} \over 2}\left\{ {\cos {{3x} \over 2} + \cos {x \over 2}} \right\} = 0
⇒2sin⁡5x2{2cos⁡xcos⁡x2}=0\Rightarrow 2\sin {{5x} \over 2}\left\{ {2\cos x\cos {x \over 2}} \right\} = 0
2sin⁡5x2=0⇒5x2=0,π,2π,3π,4π,5π2\sin {{5x} \over 2} = 0 \Rightarrow {{5x} \over 2} = 0,\pi ,2\pi ,3\pi ,4\pi ,5\pi
⇒x=0,2π5,4π5,6π5,8π5,2π\Rightarrow x = 0,{{2\pi } \over 5},{{4\pi } \over 5},{{6\pi } \over 5},{{8\pi } \over 5},2\pi
cos⁡x2=0⇒x2=π2⇒x=π\cos {x \over 2} = 0 \Rightarrow {x \over 2} = {\pi \over 2} \Rightarrow x = \pi
cos⁡x=0⇒x=π2,3π2\cos x = 0 \Rightarrow x = {\pi \over 2},{{3\pi } \over 2}

So, sum

=6π+π+2π=9π= 6\pi + \pi + 2\pi = 9\pi
Q19
The sum of solutions of the equation cos⁡x1+sin⁡x=∣tan⁡2x∣{{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right|, x∈(−π2,π2)−{π4,−π4}x \in \left( { - {\pi \over 2},{\pi \over 2}} \right) - \left\{ {{\pi \over 4}, - {\pi \over 4}} \right\} is :
A −11π30 - {{11\pi } \over {30}}
B π10{\pi \over {10}}
C −7π30 - {{7\pi } \over {30}}
D −π15 - {\pi \over {15}}
Correct Answer
Option A
Solution
cos⁡x1+sin⁡x=∣tan⁡2x∣{{\cos x} \over {1 + \sin x}} = \left| {\tan 2x} \right|
⇒cos⁡2x/2−sin⁡2x/2(cos⁡x/2+sin⁡x/2)2=∣tan⁡2x∣\Rightarrow {{{{\cos }^2}x/2 - {{\sin }^2}x/2} \over {(\cos x/2 + \sin x/2)^2}} = \left| {\tan 2x} \right|

⇒\Rightarrow

cos⁡x2−sin⁡x2cos⁡x2+sin⁡x2=∣tan⁡2x∣{{\cos {x \over 2} - \sin {x \over 2}} \over {\cos {x \over 2} + \sin {x \over 2}}} = \left| {\tan 2x} \right|

⇒\Rightarrow

1−tan⁡x21+tan⁡x2=∣tan⁡2x∣{{1 - \tan {x \over 2}} \over {1 + \tan {x \over 2}}} = \left| {\tan 2x} \right|

⇒\Rightarrow

tan⁡π4−tan⁡x2tan⁡π4+tan⁡x2=∣tan⁡2x∣{{\tan {\pi \over 4} - \tan {x \over 2}} \over {\tan {\pi \over 4} + \tan {x \over 2}}} = \left| {\tan 2x} \right|
⇒tan⁡2(π4−π2)=tan⁡22x\Rightarrow {\tan ^2}\left( {{\pi \over 4} - {\pi \over 2}} \right) = {\tan ^2}2x
⇒2x=nπ±(π4−π2)\Rightarrow 2x = n\pi \pm \left( {{\pi \over 4} - {\pi \over 2}} \right)
⇒x=−3π10,−π6,π10\Rightarrow x = {{ - 3\pi } \over {10}},{{ - \pi } \over 6},{\pi \over {10}}

or sum

=−11π6= {{ - 11\pi } \over 6}

.

Q20
The number of solutions of the equation 32tan⁡2x+32sec⁡2x=81, 0≤x≤π4{32^{{{\tan }^2}x}} + {32^{{{\sec }^2}x}} = 81,\,0 \le x \le {\pi \over 4} is :
A 3
B 1
C 0
D 2
Correct Answer
Option B
Solution
(32)tan⁡2x+(32)sec⁡2x=81{(32)^{{{\tan }^2}x}} + {(32)^{{{\sec }^2}x}} = 81
⇒(32)tan⁡2x+(32)1+tan⁡2x=81\Rightarrow {(32)^{{{\tan }^2}x}} + {(32)^{1 + {{\tan }^2}x}} = 81
⇒(32)tan⁡2x=8133\Rightarrow {(32)^{{{\tan }^2}x}} = {{81} \over {33}}

taking log of base 32 both side, ⇒\Rightarrow tan2x =

log⁡32(8133){\log _{32}}\left( {{{81} \over {33}}} \right)

⇒\Rightarrow tan x =

log⁡32(8133)\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)}

As value of

log⁡32(8133)\sqrt {{{\log }_{32}}\left( {{{81} \over {33}}} \right)}

belongs to (0, 1). In interval

 0≤x≤π4\,0 \le x \le {\pi \over 4}

only one solution.

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