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Trigonometric Equations

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Q21
If n is the number of solutions of the equation 2cos⁡x(4sin⁡(π4+x)sin⁡(π4−x)−1)=1,x∈[0,π]2\cos x\left( {4\sin \left( {{\pi \over 4} + x} \right)\sin \left( {{\pi \over 4} - x} \right) - 1} \right) = 1,x \in [0,\pi ] and S is the sum of all these solutions, then the ordered pair (n, S) is :
A (3, 13π\pi / 9)
B (2, 2π\pi / 3)
C (2, 8π\pi / 9)
D (3, 5π\pi / 3)
Correct Answer
Option A
Solution
2cos⁡x(4sin⁡(π4+x)sin⁡(π4−x)−1)=12\cos x\left( {4\sin \left( {{\pi \over 4} + x} \right)\sin \left( {{\pi \over 4} - x} \right) - 1} \right) = 1
2cos⁡x(4(sin⁡2π4−sin⁡2x)−1)=12\cos x\left( {4\left( {{{\sin }^2}{\pi \over 4} - {{\sin }^2}x} \right) - 1} \right) = 1
2cos⁡x(4(12−sin⁡2x)−1)=12\cos x\left( {4\left( {{1 \over 2} - {{\sin }^2}x} \right) - 1} \right) = 1
2cos⁡x(2−4sin⁡2x−1)=12\cos x(2 - 4{\sin ^2}x - 1) = 1
2cos⁡x(1−4sin⁡2x)=12\cos x(1 - 4{\sin ^2}x) = 1
2cos⁡x(4cos⁡2x−3)=12\cos x(4{\cos ^2}x - 3) = 1
4cos⁡3x−3cos⁡x=124{\cos ^3}x - 3\cos x = {1 \over 2}
cos⁡3x=12\cos 3x = {1 \over 2}
x∈[0,π]x \in [0,\pi ]

∴\therefore

3x∈[0,3π]3x \in [0,3\pi ]
Q22
The number of solutions of the equation cos⁡(x+π3)cos⁡(π3−x)=14cos⁡22x\cos \left( {x + {\pi \over 3}} \right)\cos \left( {{\pi \over 3} - x} \right) = {1 \over 4}{\cos ^2}2x, x∈[−3π,3π]x \in [ - 3\pi ,3\pi ] is :
A 8
B 5
C 6
D 7
Correct Answer
Option D
Solution
cos⁡(x+π3)cos⁡(π3−x)=14cos⁡22x, x∈[−3π, 3π]\cos \left( {x + {\pi \over 3}} \right)\cos \left( {{\pi \over 3} - x} \right) = {1 \over 4}{\cos ^2}2x,\,x \in [ - 3\pi ,\,3\pi ]
⇒cos⁡2x+cos⁡2π3=12cos⁡22x\Rightarrow \cos 2x + \cos {{2\pi } \over 3} = {1 \over 2}{\cos ^2}2x
⇒cos⁡22x−2cos⁡2x−1=0\Rightarrow {\cos ^2}2x - 2\cos 2x - 1 = 0
⇒cos⁡2x=1\Rightarrow \cos 2x = 1

∴\therefore

x=−3π, −2π, −π, 0, π, 2π, 3πx = - 3\pi ,\, - 2\pi ,\, - \pi ,\,0,\,\pi ,\,2\pi ,\,3\pi

∴\therefore Number of solutions = 7

Q23
Let S={θ∈[−π,π]−{±  π2}:sin⁡θtan⁡θ+tan⁡θ=sin⁡2θ}S = \left\{ {\theta \in [ - \pi ,\pi ] - \left\{ { \pm \,\,{\pi \over 2}} \right\}:\sin \theta \tan \theta + \tan \theta = \sin 2\theta } \right\}. If T=∑θ ∈ Scos⁡2θT = \sum\limits_{\theta \, \in \,S}^{} {\cos 2\theta } , then T + n(S) is equal to :
A 7 + 3\sqrt 3
B 9
C 8 + 3\sqrt 3
D 10
Correct Answer
Option B
Solution

tan⁡θ(sin⁡θ+1)−sin⁡2θ=0\tan \theta(\sin \theta+1)-\sin 2 \theta=0

tan⁡θ(sin⁡θ+1−2cos⁡2θ)=0⇒tan⁡θ=0 or 2sin⁡2θ+sin⁡θ−1=0⇒(2sin⁡θ+1)(sin⁡θ−1)=0⇒sin⁡θ=−12 or 1\begin{aligned} &\tan \theta\left(\sin \theta+1-2 \cos ^{2} \theta\right)=0 \\\\ &\Rightarrow \tan \theta=0 \text{ or } 2 \sin ^{2} \theta+\sin \theta-1=0 \\\\ &\Rightarrow(2 \sin \theta+1)(\sin \theta-1)=0 \\\\ &\Rightarrow \sin \theta=\frac{-1}{2} \text{ or } 1 \end{aligned}

But, sin⁡θ=1\sin \theta=1 not possible

θ=0,π,−π,−π6,−5π6\theta=0, \pi,-\pi,-\frac{\pi}{6}, \frac{-5 \pi}{6}
n(S)=5\mathrm{n}(\mathrm{S})=5
T=∑cos⁡2θ=cos⁡0∘+cos⁡2π+cos⁡(−2π)+cos⁡(−5π3)+cos⁡(−π3)T=\sum \cos 2 \theta=\cos 0^{\circ}+\cos 2 \pi+\cos (-2 \pi) +\cos \left(-\frac{5 \pi}{3}\right)+\cos \left(-\frac{\pi}{3}\right)

= 4

Q24
The number of solutions of ∣cos⁡x∣=sin⁡x|\cos x|=\sin x, such that −4π≤x≤4π-4 \pi \leq x \leq 4 \pi is :
A 4
B 6
C 8
D 12
Correct Answer
Option C
Solution

Period of

∣cos⁡x∣=π|\cos x| = \pi

And period of

sin⁡x=2π\sin x = 2\pi

Graph of

sin⁡x\sin x

and

∣cos⁡x∣|\cos x|

cuts each other at two points A and B in

[0,2π]\left[ {0,2\pi } \right]

So, in

[−4π,4π]\left[ { - 4\pi ,4\pi } \right]

, total 4 similar graph will be present and graph of

sin⁡x\sin x

and

∣cos⁡x∣|\cos x|

will cut

4×2=84 \times 2 = 8

times. ∴\therefore Total possible solutions = 8.

Q25
Let S={θ∈[0,2π]:82sin⁡2θ+82cos⁡2θ=16}.S=\left\{\theta \in[0,2 \pi]: 8^{2 \sin ^{2} \theta}+8^{2 \cos ^{2} \theta}=16\right\} . Then n(s)+∑θ∈S(sec⁡(π4+2θ)cos⁡ec(π4+2θ))n(s) + \sum\limits_{\theta \in S}^{} {\left( {\sec \left( {{\pi \over 4} + 2\theta } \right)\cos ec\left( {{\pi \over 4} + 2\theta } \right)} \right)} is equal to:
A 0
B −-2
C −-4
D 12
Correct Answer
Option C
Solution
S={θ∈[0,2π]:82sin⁡2θ+82cos⁡2θ=16}S = \left\{ {\theta \in [0,2\pi ]:{8^{2{{\sin }^2}\theta }} + {8^{2{{\cos }^2}\theta }} = 16} \right\}

Now apply AM ≥\ge GM for

82sin⁡2θ, 82cos⁡2θ{8^{2{{\sin }^2}\theta }},\,{8^{2{{\cos }^2}\theta }}
82sin⁡2θ+82cos⁡2θ2≥(82sin⁡2θ+2cos⁡2θ)12{{{8^{2{{\sin }^2}\theta }} + {8^{2{{\cos }^2}\theta }}} \over 2} \ge {\left( {{8^{2{{\sin }^2}\theta + 2{{\cos }^2}\theta }}} \right)^{{1 \over 2}}}
8≥88 \ge 8
⇒82sin⁡2θ=82cos⁡2θ\Rightarrow {8^{2{{\sin }^2}\theta }} = {8^{2{{\cos }^2}\theta }}

or

sin⁡2θ=cos⁡2θ{\sin ^2}\theta = {\cos ^2}\theta

∴\therefore

θ=π4,3π4,5π4,7π4\theta = {\pi \over 4},{{3\pi } \over 4},{{5\pi } \over 4},{{7\pi } \over 4}
n(S)+∑θ∈Ssec⁡(π4+2θ)cos⁡ec(π4+2θ)n(S) + \sum\limits_{\theta \in S}^{} {\sec \left( {{\pi \over 4} + 2\theta } \right)\cos ec\left( {{\pi \over 4} + 2\theta } \right)}
4+∑θ∈S22sin⁡(π4+2θ)cos⁡(π4+2θ)4 + \sum\limits_{\theta \in S}^{} {{2 \over {2\sin \left( {{\pi \over 4} + 2\theta } \right)\cos \left( {{\pi \over 4} + 2\theta } \right)}}}
=4+∑θ∈S2sin⁡(π2+4θ)=4+2∑θ∈Scos⁡ec(π2+4θ)= 4 + \sum\limits_{\theta \in S}^{} {{2 \over {\sin \left( {{\pi \over 2} + 4\theta } \right)}} = 4 + 2\sum\limits_{\theta \in S}^{} {\cos ec\left( {{\pi \over 2} + 4\theta } \right)} }
=4+2[cos⁡ec(π2+π)cos⁡ec(π2+3π)+cos⁡ec(π2+5π)+cos⁡ec(π2+7π)]= 4 + 2\left[ {\cos ec\left( {{\pi \over 2} + \pi } \right)\cos ec\left( {{\pi \over 2} + 3\pi } \right) + \cos ec\left( {{\pi \over 2} + 5\pi } \right) + \cos ec\left( {{\pi \over 2} + 7\pi } \right)} \right]
=4+2[−cos⁡ecπ2−cos⁡ecπ2−cos⁡ecπ2−cos⁡ecπ2]= 4 + 2\left[ { - \cos ec{\pi \over 2} - \cos ec{\pi \over 2} - \cos ec{\pi \over 2} - \cos ec{\pi \over 2}} \right]
=4−2(4)= 4 - 2(4)
=4−8= 4 - 8
=−4= - 4
Q26
Let S={θ∈(0,π2):∑m=19sec⁡(θ+(m−1)π6)sec⁡(θ+mπ6)=−83}S=\left\{\theta \in\left(0, \frac{\pi}{2}\right): \sum\limits_{m=1}^{9} \sec \left(\theta+(m-1) \frac{\pi}{6}\right) \sec \left(\theta+\frac{m \pi}{6}\right)=-\frac{8}{\sqrt{3}}\right\}. Then
A S={π12} S=\left\{\dfrac{\pi}{12}\right\}
B S={2π3} S=\left\{\dfrac{2 \pi}{3}\right\}
C ∑θ∈Sθ=π2 \sum\limits_{\theta \in S} \theta=\dfrac{\pi}{2}
D ∑θ∈Sθ=3π4 \sum\limits_{\theta \in S} \theta=\dfrac{3\pi}{4}
Correct Answer
Option C
Solution
s={0∈(0,π2):∑m=19sec⁡(θ+(m−1)π6)sec⁡(θ+mπ6)=−83}s = \left\{ {0 \in \left( {0,{\pi \over 2}} \right):\sum\limits_{m = 1}^9 {\sec \left( {\theta + (m - 1){\pi \over 6}} \right)\sec \left( {\theta + {{m\pi } \over 6}} \right) = - {8 \over {\sqrt 3 }}} } \right\}

.

∑m=191cos⁡(θ+(m−1)π6)cos⁡(θ+mπ6)\sum\limits_{m = 1}^9 {{1 \over {\cos \left( {\theta + (m - 1){\pi \over 6}} \right)}}\cos \left( {\theta + m{\pi \over 6}} \right)}
1sin⁡(π6)∑m=19sin⁡[(θ+mπ6)−(θ+(m−1)π6)]cos⁡(θ+(m−1)π6)cos⁡(θ+mπ6){1 \over {\sin \left( {{\pi \over 6}} \right)}}\sum\limits_{m = 1}^9 {{{\sin \left[ {\left( {\theta + {{m\pi } \over 6}} \right) - \left( {\theta + (m - 1){\pi \over 6}} \right)} \right]} \over {\cos \left( {\theta + (m - 1){\pi \over 6}} \right)\cos \left( {\theta + m{\pi \over 6}} \right)}}}
=2∑m=19[tan⁡(θ+mπ6)−tan⁡(θ+(m−1)π6)]= 2\sum\limits_{m = 1}^9 {\left[ {\tan \left( {\theta + {{m\pi } \over 6}} \right) - \tan \left( {\theta + (m - 1){\pi \over 6}} \right)} \right]}

Now,

m=12[tan⁡(θ+π6)−tan⁡(θ)]m=22[tan⁡(θ+2π6)−tan⁡(θ+π6)]...m=92[tan⁡(θ+9π6)−tan⁡(θ+8π6)]\begin{array}{ll}{m = 1} & {2\left[ {\tan \left( {\theta + {\pi \over 6}} \right) - \tan (\theta )} \right]} \\ {m = 2} & {2\left[ {\tan \left( {\theta + {{2\pi } \over 6}} \right) - \tan \left( {\theta + {\pi \over 6}} \right)} \right]} \\ \begin{array}{ll}. \\ . \\ . \end{array} & {} \\ {m = 9} & {2\left[ {\tan \left( {\theta + {{9\pi } \over 6}} \right) - \tan \left( {\theta + 8{\pi \over 6}} \right)} \right]} \end{array}

∴\therefore

=2[tan⁡(θ+3π2)−tan⁡θ]=−83= 2\left[ {\tan \left( {\theta + {{3\pi } \over 2}} \right) - \tan \theta } \right] = {{ - 8} \over {\sqrt 3 }}
=−2[cot⁡θ+tan⁡θ]=−83= - 2[\cot \theta + \tan \theta ] = {{ - 8} \over {\sqrt 3 }}
=−2×22sin⁡θcos⁡θ=−83= - {{2 \times 2} \over {2\sin \theta \cos \theta }} = {{ - 8} \over {\sqrt 3 }}
=1sin⁡2θ=23= {1 \over {\sin 2\theta }} = {2 \over {\sqrt 3 }}
⇒sin⁡2θ=32\Rightarrow \sin 2\theta = {{\sqrt 3 } \over 2}
2θ=π32\theta = {\pi \over 3}
2θ=2π32\theta = {{2\pi } \over 3}
θ=π6\theta = {\pi \over 6}
θ=π3\theta = {\pi \over 3}
∑θi=π6+π3=π2\sum {{\theta _i} = {\pi \over 6} + {\pi \over 3} = {\pi \over 2}}
Q27
The sum of the solutions x∈Rx \in \mathbb{R} of the equation 3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6\dfrac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 is
A 3
B 1
C 0
D −-1
Correct Answer
Option D
Solution
3cos⁡2x+cos⁡32xcos⁡6x−sin⁡6x=x3−x2+6⇒cos⁡2x(3+cos⁡22x)cos⁡2x(1−sin⁡2xcos⁡2x)=x3−x2+6⇒4(3+cos⁡22x)(4−sin⁡22x)=x3−x2+6⇒4(3+cos⁡22x)(3+cos⁡22x)=x3−x2+6x3−x2+2=0⇒(x+1)(x2−2x+2)=0\begin{aligned} & \frac{3 \cos 2 x+\cos ^3 2 x}{\cos ^6 x-\sin ^6 x}=x^3-x^2+6 \\ & \Rightarrow \frac{\cos 2 x\left(3+\cos ^2 2 x\right)}{\cos 2 x\left(1-\sin ^2 x \cos ^2 x\right)}=x^3-x^2+6 \\ & \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(4-\sin ^2 2 x\right)}=x^3-x^2+6 \\ & \Rightarrow \frac{4\left(3+\cos ^2 2 x\right)}{\left(3+\cos ^2 2 x\right)}=x^3-x^2+6 \\ & x^3-x^2+2=0 \Rightarrow(x+1)\left(x^2-2 x+2\right)=0 \end{aligned}

so, sum of real solutions

=−1=-1
Q28
If 2sin⁡3x+sin⁡2xcos⁡x+4sin⁡x−4=02 \sin ^3 x+\sin 2 x \cos x+4 \sin x-4=0 has exactly 3 solutions in the interval [0,nπ2],n∈N\left[0, \dfrac{\mathrm{n} \pi}{2}\right], \mathrm{n} \in \mathrm{N}, then the roots of the equation x2+nx+(n−3)=0x^2+\mathrm{n} x+(\mathrm{n}-3)=0 belong to :
A (0,∞)(0, \infty)
B Z
C (−172,172)\left(-\dfrac{\sqrt{17}}{2}, \dfrac{\sqrt{17}}{2}\right)
D (−∞,0)(-\infty, 0)
Correct Answer
Option D
Solution
2sin⁡3x+2sin⁡x⋅cos⁡2x+4sin⁡x−4=02sin⁡3x+2sin⁡x⋅(1−sin⁡2x)+4sin⁡x−4=06sin⁡x−4=0sin⁡x=23n=5 (in the given interval) x2+5x+2=0x=−5±172 Required interval (−∞,0)\begin{aligned} & 2 \sin ^3 x+2 \sin x \cdot \cos ^2 x+4 \sin x-4=0 \\ & 2 \sin ^3 x+2 \sin x \cdot\left(1-\sin ^2 x\right)+4 \sin x-4=0 \\ & 6 \sin x-4=0 \\ & \sin x=\frac{2}{3} \\ & \mathbf{n}=5 \text{ (in the given interval) } \\ & x^2+5 x+2=0 \\ & x=\frac{-5 \pm \sqrt{17}}{2} \\ & \text{ Required interval }(-\infty, 0) \end{aligned}
Q29
The number of solutions of the equation cos⁡2θcos⁡θ2+cos⁡5θ2=2cos⁡35θ2 \cos 2\theta \cos \dfrac{\theta}{2} + \cos \dfrac{5\theta}{2} = 2\cos^3 \dfrac{5\theta}{2} in [−π2,π2] \left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right] is :
A 5
B 7
C 6
D 9
Correct Answer
Option B
Solution
cos⁡2θcos⁡θ2+cos⁡5θ2=2cos⁡35θ212(2cos⁡2θcos⁡θ2)+cos⁡502=12(cos⁡15θ2+3cos⁡5θ2) or solving cos⁡3θ2=cos⁡15θ2cos⁡15θ2−cos⁡3θ2=02sin⁡30sin⁡9θ2=03θ=nπ or 9θ2=mπθ=nπ3θ=2 mπ9θ={−π2,π3,0}θ={−4π9,−2π9,4π9,2π9}\begin{aligned} &\begin{aligned} & \cos 2 \theta \cos \frac{\theta}{2}+\cos \frac{5 \theta}{2}=2 \cos ^3 \frac{5 \theta}{2} \\ & \frac{1}{2}\left(2 \cos 2 \theta \cos \frac{\theta}{2}\right)+\cos \frac{50}{2} \\ & =\frac{1}{2}\left(\cos \frac{15 \theta}{2}+3 \cos \frac{5 \theta}{2}\right) \end{aligned}\\ &\text{ or solving }\\ &\begin{aligned} & \cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2} \\ & \cos \frac{15 \theta}{2}-\cos \frac{3 \theta}{2}=0 \\ & 2 \sin 30 \sin \frac{9 \theta}{2}=0 \end{aligned}\\ &3 \theta=\mathrm{n} \pi \quad \text{ or } \frac{9 \theta}{2}=\mathrm{m} \pi\\ &\begin{aligned} & \theta=\frac{\mathrm{n} \pi}{3} \quad \theta=\frac{2 \mathrm{~m} \pi}{9} \\ & \theta=\left\{-\frac{\pi}{2}, \frac{\pi}{3}, 0\right\} \\ & \theta=\left\{-\frac{4 \pi}{9}, \frac{-2 \pi}{9}, \frac{4 \pi}{9}, \frac{2 \pi}{9}\right\} \end{aligned} \end{aligned}
Q30
The number of elements in the set S={θ∈[0,2π]:3cos⁡4θ−5cos⁡2θ−2sin⁡6θ+2=0}S=\left\{\theta \in[0,2 \pi]: 3 \cos ^{4} \theta-5 \cos ^{2} \theta-2 \sin ^{6} \theta+2=0\right\} is :
A 9
B 8
C 12
D 10
Correct Answer
Option A
Solution

The original equation is:

3cos⁡4θ−5cos⁡2θ−2sin⁡6θ+2=03 \cos ^4 \theta-5 \cos ^2 \theta-2 \sin ^6 \theta+2=0

We can re-arrange terms and use trigonometric identity (cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1) to rewrite this as :

3cos⁡4θ−3cos⁡2θ+2sin⁡2θ−2sin⁡6θ=03 \cos ^4 \theta - 3 \cos ^2 \theta + 2 \sin ^2 \theta - 2 \sin ^6 \theta = 0

Factoring out cos⁡2θ\cos^2 \theta and sin⁡2θ\sin^2 \theta we get :

3cos⁡2θ(cos⁡2θ−1)+2sin⁡2θ(sin⁡4θ−1)=03 \cos ^2 \theta (\cos ^2 \theta - 1) + 2 \sin ^2 \theta (\sin ^4 \theta - 1) = 0

Then we substitute 1−cos⁡2θ1 - \cos^2 \theta for sin⁡2θ\sin^2 \theta (again using the same trigonometric identity), which simplifies to :

−3cos⁡2θsin⁡2θ+2sin⁡2θ(1+sin⁡2θ)cos⁡2θ=0-3 \cos ^2 \theta \sin ^2 \theta + 2 \sin ^2 \theta(1 + \sin ^2 \theta)\cos ^2 \theta = 0

This can be re-written as :

sin⁡2θcos⁡2θ(2+2sin⁡2θ−3)=0\sin ^2 \theta \cos ^2 \theta(2 + 2 \sin ^2 \theta - 3) = 0

Simplifying it to :

sin⁡2θcos⁡2θ(2sin⁡2θ−1)=0\sin ^2 \theta \cos ^2 \theta(2 \sin ^2 \theta - 1) = 0

We now have 3 separate conditions to solve for : - sin⁡2θ=0\sin ^2 \theta = 0, - cos⁡2θ=0\cos ^2 \theta = 0, and - 2sin⁡2θ−1=02 \sin ^2 \theta - 1 = 0.

For sin⁡2θ=0\sin ^2 \theta = 0, the solutions are θ={0,π,2π}\theta=\{0, \pi, 2 \pi\} (3 solutions).

For cos⁡2θ=0\cos ^2 \theta = 0, the solutions are θ={π2,3π2}\theta=\left\{\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\right\} (2 solutions).

For 2sin⁡2θ−1=02 \sin ^2 \theta - 1 = 0 (which simplifies to sin⁡2θ=1/2\sin ^2 \theta = 1/2), the solutions are θ={π4,3π4,5π4,7π4}\theta=\left\{\dfrac{\pi}{4}, \dfrac{3 \pi}{4}, \dfrac{5 \pi}{4}, \dfrac{7 \pi}{4}\right\} (4 solutions).

In total, this gives 3+2+4 = 9 solutions.

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