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Trigonometric Equations

JEE Mathematics · 41 questions · Page 4 of 5 · Click an option or "Show Solution" to reveal answer

Q31
Let S={x∈(−π2,π2):91−tan⁡2x+9tan⁡2x=10}S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\} and β=∑x∈Stan⁡2(x3)\beta=\sum\limits_{x \in S} \tan ^{2}\left(\dfrac{x}{3}\right), then 16(β−14)2\dfrac{1}{6}(\beta-14)^{2} is equal to :
A 16
B 32
C 8
D 64
Correct Answer
Option B
Solution

We have,

S={x∈(−π2,π2):91−tan⁡2x+9tan⁡2x=10}S=\left\{x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): 9^{1-\tan ^{2} x}+9^{\tan ^{2} x}=10\right\}

and

β=∑x∈Stan⁡2(x3)\beta=\sum\limits_{x \in S} \tan ^{2}\left(\frac{x}{3}\right)
91−tan⁡2x+9tan⁡2x=10⇒99tan⁡2x+9tan⁡2x=10\begin{aligned} & 9^{1-\tan ^2 x}+9^{\tan ^2 x}=10 \\\\ &\Rightarrow \frac{9}{9^{\tan ^2 x}}+9^{\tan ^2 x}=10 \end{aligned}

Let 9tan⁡2x=t9^{\tan ^2 x}=t

9t+t=10⇒t2−10t+9=0⇒t=9 or 1\frac{9}{t}+t=10 \Rightarrow t^2-10 t+9=0 \Rightarrow t=9 \text{ or } 1

If t=9,9tan⁡2x=9⇒x=±π4t=9,9^{\tan ^2 x}=9 \Rightarrow x= \pm \dfrac{\pi}{4} If t=1,9tan⁡2x=1⇒x=0t=1,9^{\tan ^2 x}=1 \Rightarrow x=0 Also,

β=∑x∈Stan⁡2(x3)=tan⁡2(03)+tan⁡2(π12)+tan⁡2(−π12)=0+2tan⁡2π12=2[2−3]2=2[4+3−43]=14−83\begin{aligned} \beta & =\sum_{x \in S} \tan ^2\left(\frac{x}{3}\right) \\\\ & =\tan ^2\left(\frac{0}{3}\right)+\tan ^2\left(\frac{\pi}{12}\right)+\tan ^2\left(-\frac{\pi}{12}\right) \\\\ & =0+2 \tan ^2 \frac{\pi}{12}=2[2-\sqrt{3}]^2 \\\\ & =2[4+3-4 \sqrt{3}]=14-8 \sqrt{3} \end{aligned}
 So, now 16(β−14)2=16[14−83−14]2=16×64×3=32\text{ So, now } \frac{1}{6}(\beta-14)^2=\frac{1}{6}[14-8 \sqrt{3}-14]^2=\frac{1}{6} \times 64 \times 3=32
Q32
The number of solutions of the equation 4sin⁡2x−4cos⁡3x+9−4cos⁡x=0;x∈[−2π,2π]4 \sin ^2 x-4 \cos ^3 x+9-4 \cos x=0 ; x \in[-2 \pi, 2 \pi] is :
A 0
B 3
C 1
D 2
Correct Answer
Option A
Solution

We start by recognizing that sin⁡2x+cos⁡2x=1 \sin^2 x + \cos^2 x = 1.

Substituting sin⁡2x=1−cos⁡2x \sin^2 x = 1 - \cos^2 x into the original equation gives:

4(1−cos⁡2x)−4cos⁡3x+9−4cos⁡x=04(1 - \cos^2 x) - 4 \cos^3 x + 9 - 4 \cos x = 0

Rearranging and simplifying this equation, we have:

4−4cos⁡2x−4cos⁡3x+9−4cos⁡x=04 - 4 \cos^2 x - 4 \cos^3 x + 9 - 4 \cos x = 0
4cos⁡3x+4cos⁡2x+4cos⁡x−13=04 \cos^3 x + 4 \cos^2 x + 4 \cos x - 13 = 0

4cos⁡3x+4cos⁡2x+4cos⁡x=134 \cos ^3 x+4 \cos ^2 x+4 \cos x=13 Observing the bounds given, x∈[−2π,2π]x \in [-2\pi, 2\pi], we want to find how many solutions satisfy this cubic equation in terms of cos⁡x \cos x.

However, we note that the left-hand side (LHS) of the equation, representing a combination of cosines, could at most approach a maximum sum when cos⁡x=1 \cos x = 1, that being 4(1)+4(1)+4(1)=124(1) + 4(1) + 4(1) = 12.

Yet, we have the equation set to equal 13, which is impossible given the maximum sum of the LHS can only be 12.

The above reasoning indicates that, within the domain specified, there is no value of xx for which the equation holds true.

Therefore, the number of solutions to the equation is zero.

Q33
If 2tan⁡2θ−5sec⁡θ=12 \tan ^2 \theta-5 \sec \theta=1 has exactly 7 solutions in the interval [0,nπ2]\left[0, \dfrac{n \pi}{2}\right], for the least value of n∈Nn \in \mathbf{N}, then ∑k=1nk2k\sum\limits_{k=1}^n \dfrac{k}{2^k} is equal to:
A 1214(215−15)\dfrac{1}{2^{14}}\left(2^{15}-15\right)
B 1−152131-\dfrac{15}{2^{13}}
C 1215(214−14)\dfrac{1}{2^{15}}\left(2^{14}-14\right)
D 1213(214−15)\dfrac{1}{2^{13}}\left(2^{14}-15\right)
Correct Answer
Option D
Solution
2tan⁡2θ−5sec⁡θ−1=0⇒2sec⁡2θ−5sec⁡θ−3=0⇒(2sec⁡θ+1)(sec⁡θ−3)=0⇒sec⁡θ=−12,3⇒cos⁡θ=−2,13⇒cos⁡θ=13\begin{aligned} & 2 \tan ^2 \theta-5 \sec \theta-1=0 \\ & \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 \\ & \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \\ & \Rightarrow \sec \theta=-\frac{1}{2}, 3 \\ & \Rightarrow \cos \theta=-2, \frac{1}{3} \\ & \Rightarrow \cos \theta=\frac{1}{3} \end{aligned}

For 7 solutions

n=13\mathrm{n}=13
 So, ∑k=113k2k=S (say) S=12+222+323+….+1321312 S=122+123+….+12213+13214⇒S2=12⋅1−12131−12−13214⇒S=2⋅(213−1213)−13213\begin{aligned} & \text{ So, } \sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S} \text{ (say) } \\ & \mathrm{S}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots .+\frac{13}{2^{13}} \\ & \frac{1}{2} \mathrm{~S}=\frac{1}{2^2}+\frac{1}{2^3}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}} \\ & \Rightarrow \frac{\mathrm{S}}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow \mathrm{S}=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}} \end{aligned}
Q34
Let ∣cos⁡θcos⁡(60−θ)cos⁡(60+θ)∣≤18,θϵ[0,2π]|\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \dfrac{1}{8}, \theta \epsilon[0,2 \pi]. Then, the sum of all θ∈[0,2π]\theta \in[0,2 \pi], where cos⁡3θ\cos 3 \theta attains its maximum value, is :
A 6π6 \pi
B 9π9 \pi
C 18π18 \pi
D 15π15 \pi
Correct Answer
Option A
Solution
∣cos⁡θcos⁡(60−θ)cos⁡(60+θ)∣≤18⇒14∣cos⁡3θ∣≤18cos⁡3θ is max if cos⁡3θ=12∴θ=π9,5π9,7π9,11π9,13π9,17π9∑θi=6π\begin{aligned} & |\cos \theta \cos (60-\theta) \cos (60+\theta)| \leq \frac{1}{8} \\ & \Rightarrow \frac{1}{4}|\cos 3 \theta| \leq \frac{1}{8} \\ & \cos 3 \theta \text{ is max if } \cos 3 \theta=\frac{1}{2} \\ & \therefore \theta=\frac{\pi}{9}, \frac{5 \pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9} \\ & \sum \theta_i=6 \pi \end{aligned}
Q35
The sum of all values of θ∈[0,2π]\theta \in[0,2 \pi] satisfying 2sin⁡2θ=cos⁡2θ2 \sin ^2 \theta=\cos 2 \theta and 2cos⁡2θ=3sin⁡θ2 \cos ^2 \theta=3 \sin \theta is
A π\pi
B 5π6\dfrac{5 \pi}{6}
C π2\dfrac{\pi}{2}
D 4π4 \pi
Correct Answer
Option A
Solution
2sin⁡2θ=cos⁡2θ2sin⁡2θ=1−2sin⁡2θ4sin⁡2θ=1sin⁡2θ=14sin⁡θ=±122cos⁡2θ=3sin⁡θ2−2sin⁡2θ+3sin⁡θ−2=0(2sin⁡θ−1)(2sin⁡θ−2)=0sin⁡θ=12\begin{aligned} & 2 \sin ^2 \theta=\cos 2 \theta \\ & 2 \sin ^2 \theta=1-2 \sin ^2 \theta \\ & 4 \sin ^2 \theta=1 \\ & \sin ^2 \theta=\frac{1}{4} \\ & \sin \theta= \pm \frac{1}{2} \\ & 2 \cos ^2 \theta=3 \sin \theta \\ & 2-2 \sin ^2 \theta+3 \sin \theta-2=0 \\ & (2 \sin \theta-1)(2 \sin \theta-2)=0 \\ & \sin \theta=\frac{1}{2} \end{aligned}

so common equation which satisfy both equations is sin⁡θ=12\sin \theta=\dfrac{1}{2}

θ=π6,5π6(θ∈[0,2π])\theta=\frac{\pi}{6}, \frac{5 \pi}{6} \quad(\theta \in[0,2 \pi])
 Sum =π\text{ Sum }=\pi
Q36
If θ∈[−2π,2π]\theta \in[-2 \pi, 2 \pi], then the number of solutions of 22cos⁡2θ+(2−6)cos⁡θ−3=02 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0, is equal to:
A 8
B 6
C 10
D 12
Correct Answer
Option A
Solution
22cos⁡2θ+(2−6)cos⁡θ−3=022cos⁡2θ+2cos⁡θ−6cos⁡θ−3=0(2cos⁡θ−3)(2cos⁡θ+1)=0⇒cos⁡θ=32 or cos⁡θ=−12θ={−11π6,−5π4,−3π4,−π6,π6,3π4,5π4,11π6}⇒8 (solution) \begin{aligned} & 2 \sqrt{2} \cos ^2 \theta+(2-\sqrt{6}) \cos \theta-\sqrt{3}=0 \\ & 2 \sqrt{2} \cos ^2 \theta+2 \cos \theta-\sqrt{6} \cos \theta-\sqrt{3}=0 \\ & (2 \cos \theta-\sqrt{3})(\sqrt{2} \cos \theta+1)=0 \\ & \Rightarrow \cos \theta=\frac{\sqrt{3}}{2} \text{ or } \cos \theta=\frac{-1}{\sqrt{2}} \\ & \theta=\left\{\frac{-11 \pi}{6}, \frac{-5 \pi}{4}, \frac{-3 \pi}{4}, \frac{-\pi}{6}, \frac{\pi}{6}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{11 \pi}{6}\right\} \\ & \Rightarrow 8 \text{ (solution) } \end{aligned}
Q37
If θϵ[−7π6,4π3]\theta \epsilon\left[-\dfrac{7 \pi}{6}, \dfrac{4 \pi}{3}\right], then the number of solutions of 3cosec⁡2θ−2(3−1)cosec⁡θ−4=0\sqrt{3} \operatorname{cosec}^2 \theta-2(\sqrt{3}-1) \operatorname{cosec} \theta-4=0, is equal to :
A 7
B 10
C 6
D 8
Correct Answer
Option C
Solution

To find the solutions of the equation 3cosec⁡2θ−2(3−1)cosec⁡θ−4=0\sqrt{3} \operatorname{cosec}^2 \theta - 2(\sqrt{3} - 1) \operatorname{cosec} \theta - 4 = 0, we can begin by solving for cosec⁡θ\operatorname{cosec} \theta.

The quadratic equation in terms of cosec⁡θ\operatorname{cosec} \theta is: 3cosec⁡2θ−2(3−1)cosec⁡θ−4=0 \sqrt{3} \operatorname{cosec}^2 \theta - 2(\sqrt{3} - 1) \operatorname{cosec} \theta - 4 = 0 Using the quadratic formula: cosec⁡θ=23−2±(2(3−1))2+43⋅423 \operatorname{cosec} \theta = \dfrac{2\sqrt{3} - 2 \pm \sqrt{(2(\sqrt{3} - 1))^2 + 4 \sqrt{3} \cdot 4}}{2\sqrt{3}} Simplifying inside the square root: (2(3−1))2+43⋅4=4(3+1−23)+163 (2(\sqrt{3} - 1))^2 + 4\sqrt{3} \cdot 4 = 4(3 + 1 - 2\sqrt{3}) + 16\sqrt{3} This simplifies to: 4(4−23)+163=16−83+163=16+83 4(4 - 2\sqrt{3}) + 16\sqrt{3} = 16 - 8\sqrt{3} + 16\sqrt{3} = 16 + 8\sqrt{3} Therefore, the quadratic gives: cosec⁡θ=23−2±2(3+1)23 \operatorname{cosec} \theta = \dfrac{2\sqrt{3} - 2 \pm 2(\sqrt{3} + 1)}{2\sqrt{3}} By solving, we find: cosec⁡θ=−23,2 \operatorname{cosec} \theta = \dfrac{-2}{\sqrt{3}}, 2 Thus, for each value of the cosecant, θ\theta can take several values that fall within the given interval [−7π6,4π3][- \dfrac{7\pi}{6}, \dfrac{4\pi}{3}]: θ=−7π6,−2π3,−π3,π6,5π6,4π3 \theta = \dfrac{-7\pi}{6}, \dfrac{-2\pi}{3}, \dfrac{-\pi}{3}, \dfrac{\pi}{6}, \dfrac{5\pi}{6}, \dfrac{4\pi}{3} These are all the possible solutions for θ\theta within the defined range.

Q38
 The number of solutions of the equation 2x+3tan⁡x=π,x∈[−2π,2π]−{±π2,±3π2} is:  \text{ The number of solutions of the equation } 2 x+3 \tan x=\pi, x \in[-2 \pi, 2 \pi]-\left\{ \pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}\right\} \text{ is: }
A 4
B 5
C 3
D 6
Correct Answer
Option B
Solution
2x+3tan⁡x=π⇒tan⁡x=π−2x3,x∈[−2π,2π]−{±π2,±3π2}\begin{aligned} & 2 x+3 \tan x=\pi \\ & \Rightarrow \quad \tan x=\frac{\pi-2 x}{3}, x \in[-2 \pi, 2 \pi]-\left\{ \pm \frac{\pi}{2}, \pm \frac{3 \pi}{2}\right\} \end{aligned}

5 solutions

Q39
The number of solutions of the equation (4−3)sin⁡x−23cos⁡2x=−41+3,x∈[−2π,5π2](4-\sqrt{3}) \sin x-2 \sqrt{3} \cos ^2 x=-\dfrac{4}{1+\sqrt{3}}, x \in\left[-2 \pi, \dfrac{5 \pi}{2}\right] is
A 4
B 3
C 6
D 5
Correct Answer
Option D
Solution

To find the number of solutions to the equation (4−3)sin⁡x−23cos⁡2x=−41+3,x∈[−2π,5π2] (4-\sqrt{3}) \sin x - 2 \sqrt{3} \cos^2 x = -\dfrac{4}{1+\sqrt{3}}, \quad x \in \left[-2\pi, \dfrac{5\pi}{2}\right] we start by letting sin⁡x=t\sin x = t, which implies cos⁡2x=1−t2\cos^2 x = 1 - t^2.

Substituting these into the equation, we have: (4−3)t−23(1−t2)=−41+3 (4-\sqrt{3}) t - 2 \sqrt{3}(1-t^2) = \dfrac{-4}{1+\sqrt{3}} Simplifying, we get: 23t2+(4−3)t−23+41+3=0 2 \sqrt{3} t^2 + (4-\sqrt{3}) t - 2\sqrt{3} + \dfrac{4}{1+\sqrt{3}} = 0 Further simplification yields: 23t2+(4−3)t+−23−21+3=0 2 \sqrt{3} t^2 + (4-\sqrt{3}) t + \dfrac{-2 \sqrt{3} - 2}{1+\sqrt{3}} = 0 This simplifies to: 23t2+(4−3)t−2=0 2 \sqrt{3} t^2 + (4-\sqrt{3}) t - 2 = 0 To find tt, solve the quadratic equation: t=(3−4)±19−83+8(23)43 t = \dfrac{(\sqrt{3}-4) \pm \sqrt{19 - 8 \sqrt{3} + 8(2\sqrt{3})}}{4 \sqrt{3}} This simplifies to: t=(3−4)±19+8343=(3−4)±(3+4)43 t = \dfrac{(\sqrt{3}-4) \pm \sqrt{19 + 8 \sqrt{3}}}{4 \sqrt{3}} = \dfrac{(\sqrt{3}-4) \pm (\sqrt{3}+4)}{4 \sqrt{3}} Evaluating further, we find: t=2343or−843 t = \dfrac{2 \sqrt{3}}{4 \sqrt{3}} \quad \text{or} \quad \dfrac{-8}{4 \sqrt{3}} This gives us sin⁡x=12 \sin x = \dfrac{1}{2} or −23Consideringtheinterval\dfrac{-2}{\sqrt{3}} Considering the interval x \in \left[-2\pi, \frac{5\pi}{2}\right],wefindthat, we find that \sin x = \frac{1}{2}$ occurs at several points.

After checking, there are 5 such solutions in the given interval.

Q40
If 0 ≤ \le x < π2{\pi \over 2}, then the number of values of x for which sin x −- sin 2x + sin 3x = 0, is :
A 3
B 1
C 4
D 2
Correct Answer
Option D
Solution

sin x −- sin 2x + sin 3x = 0

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

⇒\Rightarrow (sin3x + sinx) −- sin2x = 0 ⇒\Rightarrow 2sin2x.cos2x −- sin2x = 0 ⇒\Rightarrow sin2x (2cosx −- 1) = 0 sin 2x = 0 x = 0 and cos x =

12{1 \over 2}

and x =

π3{\pi \over 3}

two solutions

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