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Question: The total number of real solutions of the equation $\theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2}...

The total number of real solutions of the equation

θ=tan⁡−1(2tan⁡θ)−12sin⁡−1(6tan⁡θ9+tan⁡2θ)\theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2}\sin^{-1}(\frac{6 \tan \theta}{9 + \tan^2 \theta})

is

(Here, the inverse trigonometric functions sin⁡−1x\sin^{-1} x and tan⁡−1x\tan^{-1} x assume values in [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}] and (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), respectively.)

A

1

B

2

C

3

D

5

Answer

3

Explanation

Solution

Let t=tan⁡θt = \tan \theta. The given equation is θ=tan⁡−1(2t)−12sin⁡−1(6t9+t2)\theta = \tan^{-1}(2t) - \frac{1}{2}\sin^{-1}\left(\frac{6t}{9 + t^2}\right) Let t=3tan⁡ϕt = 3 \tan \phi. Then 6t9+t2=6(3tan⁡ϕ)9+(3tan⁡ϕ)2=18tan⁡ϕ9(1+tan⁡2ϕ)=2tan⁡ϕ1+tan⁡2ϕ=sin⁡(2ϕ)\frac{6t}{9 + t^2} = \frac{6(3 \tan \phi)}{9 + (3 \tan \phi)^2} = \frac{18 \tan \phi}{9(1 + \tan^2 \phi)} = \frac{2 \tan \phi}{1 + \tan^2 \phi} = \sin(2\phi). The term 12sin⁡−1(6t9+t2)\frac{1}{2}\sin^{-1}\left(\frac{6t}{9 + t^2}\right) becomes 12sin⁡−1(sin⁡(2ϕ))\frac{1}{2}\sin^{-1}(\sin(2\phi)).

We need to consider different intervals for ϕ\phi based on the range of sin⁡−1x\sin^{-1} x, which is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. sin⁡−1(sin⁡(2ϕ))=2ϕ\sin^{-1}(\sin(2\phi)) = 2\phi if 2ϕ∈[−π2,π2]2\phi \in [-\frac{\pi}{2}, \frac{\pi}{2}], i.e., ϕ∈[−π4,π4]\phi \in [-\frac{\pi}{4}, \frac{\pi}{4}]. This corresponds to t=3tan⁡ϕ∈[3tan⁡(−π4),3tan⁡(π4)]=[−3,3]t = 3 \tan \phi \in [3 \tan(-\frac{\pi}{4}), 3 \tan(\frac{\pi}{4})] = [-3, 3].

Case 1: t∈[−3,3]t \in [-3, 3]. In this case, 12sin⁡−1(6t9+t2)=12(2ϕ)=ϕ\frac{1}{2}\sin^{-1}(\frac{6t}{9 + t^2}) = \frac{1}{2}(2\phi) = \phi. Since t=3tan⁡ϕt = 3 \tan \phi, ϕ=tan⁡−1(t/3)\phi = \tan^{-1}(t/3). The equation becomes θ=tan⁡−1(2t)−tan⁡−1(t/3)\theta = \tan^{-1}(2t) - \tan^{-1}(t/3). We use the formula tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy)\tan^{-1} x - \tan^{-1} y = \tan^{-1}(\frac{x-y}{1+xy}) for xy>−1xy > -1. Here x=2t,y=t/3x=2t, y=t/3. xy=2t2/3xy = 2t^2/3. Since tt is real, t2≥0t^2 \ge 0, so 2t2/3≥0>−12t^2/3 \ge 0 > -1. The formula is valid. θ=tan⁡−1(2t−t/31+(2t)(t/3))=tan⁡−1(5t/31+2t2/3)=tan⁡−1(5t3+2t2)\theta = \tan^{-1}\left(\frac{2t - t/3}{1 + (2t)(t/3)}\right) = \tan^{-1}\left(\frac{5t/3}{1 + 2t^2/3}\right) = \tan^{-1}\left(\frac{5t}{3 + 2t^2}\right). Since t=tan⁡θt = \tan \theta, we have tan⁡θ=tan⁡(tan⁡−1(5t3+2t2))\tan \theta = \tan\left(\tan^{-1}\left(\frac{5t}{3 + 2t^2}\right)\right). This implies t=5t3+2t2t = \frac{5t}{3 + 2t^2}. t(3+2t2)=5tt(3 + 2t^2) = 5t 3t+2t3=5t3t + 2t^3 = 5t 2t3−2t=02t^3 - 2t = 0 2t(t2−1)=02t(t^2 - 1) = 0 2t(t−1)(t+1)=02t(t-1)(t+1) = 0. The possible values for tt are 0,1,−10, 1, -1. All these values are in the interval [−3,3][-3, 3].

If t=0t=0, tan⁡θ=0\tan \theta = 0. θ=nπ\theta = n\pi. Substitute into the original equation: nπ=tan⁡−1(0)−12sin⁡−1(0)=0−0=0n\pi = \tan^{-1}(0) - \frac{1}{2}\sin^{-1}(0) = 0 - 0 = 0. nπ=0  ⟹  n=0n\pi = 0 \implies n=0. So θ=0\theta = 0 is a solution.

If t=1t=1, tan⁡θ=1\tan \theta = 1. θ=nπ+π4\theta = n\pi + \frac{\pi}{4}. Substitute into the original equation: nπ+π4=tan⁡−1(2)−12sin⁡−1(610)=tan⁡−1(2)−12sin⁡−1(35)n\pi + \frac{\pi}{4} = \tan^{-1}(2) - \frac{1}{2}\sin^{-1}(\frac{6}{10}) = \tan^{-1}(2) - \frac{1}{2}\sin^{-1}(\frac{3}{5}). We know tan⁡−1(2)−12sin⁡−1(35)=tan⁡−1(2)−tan⁡−1(13)=tan⁡−1(2−1/31+2⋅1/3)=tan⁡−1(5/35/3)=tan⁡−1(1)=π4\tan^{-1}(2) - \frac{1}{2}\sin^{-1}(\frac{3}{5}) = \tan^{-1}(2) - \tan^{-1}(\frac{1}{3}) = \tan^{-1}(\frac{2 - 1/3}{1 + 2 \cdot 1/3}) = \tan^{-1}(\frac{5/3}{5/3}) = \tan^{-1}(1) = \frac{\pi}{4}. So nπ+π4=π4  ⟹  nπ=0  ⟹  n=0n\pi + \frac{\pi}{4} = \frac{\pi}{4} \implies n\pi = 0 \implies n=0. So θ=π4\theta = \frac{\pi}{4} is a solution.

If t=−1t=-1, tan⁡θ=−1\tan \theta = -1. θ=nπ−π4\theta = n\pi - \frac{\pi}{4}. Substitute into the original equation: nπ−π4=tan⁡−1(−2)−12sin⁡−1(−610)=−tan⁡−1(2)−12(−sin⁡−1(35))=−tan⁡−1(2)+12sin⁡−1(35)n\pi - \frac{\pi}{4} = \tan^{-1}(-2) - \frac{1}{2}\sin^{-1}(\frac{-6}{10}) = -\tan^{-1}(2) - \frac{1}{2}(-\sin^{-1}(\frac{3}{5})) = -\tan^{-1}(2) + \frac{1}{2}\sin^{-1}(\frac{3}{5}). We know tan⁡−1(2)−12sin⁡−1(35)=π4\tan^{-1}(2) - \frac{1}{2}\sin^{-1}(\frac{3}{5}) = \frac{\pi}{4}. So −tan⁡−1(2)+12sin⁡−1(35)=−(tan⁡−1(2)−12sin⁡−1(35))=−π4-\tan^{-1}(2) + \frac{1}{2}\sin^{-1}(\frac{3}{5}) = -(\tan^{-1}(2) - \frac{1}{2}\sin^{-1}(\frac{3}{5})) = -\frac{\pi}{4}. So nπ−π4=−π4  ⟹  nπ=0  ⟹  n=0n\pi - \frac{\pi}{4} = -\frac{\pi}{4} \implies n\pi = 0 \implies n=0. So θ=−π4\theta = -\frac{\pi}{4} is a solution. From Case 1, we have three solutions: θ=0,π4,−π4\theta = 0, \frac{\pi}{4}, -\frac{\pi}{4}.

Case 2: t>3t > 3. If t>3t > 3, then tan⁡ϕ>1\tan \phi > 1, so ϕ∈(π4+kπ,π2+kπ)\phi \in (\frac{\pi}{4} + k\pi, \frac{\pi}{2} + k\pi) for integer kk. We need to evaluate sin⁡−1(sin⁡(2ϕ))\sin^{-1}(\sin(2\phi)). The range of sin⁡−1\sin^{-1} is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}]. If k=0k=0, ϕ∈(π4,π2)\phi \in (\frac{\pi}{4}, \frac{\pi}{2}). Then 2ϕ∈(π2,π)2\phi \in (\frac{\pi}{2}, \pi). sin⁡−1(sin⁡(2ϕ))=π−2ϕ\sin^{-1}(\sin(2\phi)) = \pi - 2\phi. 12sin⁡−1(sin⁡(2ϕ))=π2−ϕ=π2−tan⁡−1(t/3)\frac{1}{2}\sin^{-1}(\sin(2\phi)) = \frac{\pi}{2} - \phi = \frac{\pi}{2} - \tan^{-1}(t/3). The equation is θ=tan⁡−1(2t)−(π2−tan⁡−1(t/3))=tan⁡−1(2t)+tan⁡−1(t/3)−π2\theta = \tan^{-1}(2t) - (\frac{\pi}{2} - \tan^{-1}(t/3)) = \tan^{-1}(2t) + \tan^{-1}(t/3) - \frac{\pi}{2}. Since t>3t > 3, 2t>62t > 6 and t/3>1t/3 > 1. x=2t>0,y=t/3>0x=2t > 0, y=t/3 > 0. xy=2t2/3>2(32)/3=6>1xy = 2t^2/3 > 2(3^2)/3 = 6 > 1. Use tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy)\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}(\frac{x+y}{1-xy}) for x>0,y>0,xy>1x>0, y>0, xy>1. tan⁡−1(2t)+tan⁡−1(t/3)=π+tan⁡−1(2t+t/31−(2t)(t/3))=π+tan⁡−1(7t/31−2t2/3)=π+tan⁡−1(7t3−2t2)\tan^{-1}(2t) + \tan^{-1}(t/3) = \pi + \tan^{-1}(\frac{2t + t/3}{1 - (2t)(t/3)}) = \pi + \tan^{-1}(\frac{7t/3}{1 - 2t^2/3}) = \pi + \tan^{-1}(\frac{7t}{3 - 2t^2}). The equation is θ=π+tan⁡−1(7t3−2t2)−π2=π2+tan⁡−1(7t3−2t2)\theta = \pi + \tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{\pi}{2} = \frac{\pi}{2} + \tan^{-1}(\frac{7t}{3 - 2t^2}). Since t=tan⁡θt = \tan \theta, tan⁡θ=tan⁡(π2+tan⁡−1(7t3−2t2))=−cot⁡(tan⁡−1(7t3−2t2))=−1tan⁡(tan⁡−1(7t3−2t2))=−17t3−2t2=2t2−37t\tan \theta = \tan(\frac{\pi}{2} + \tan^{-1}(\frac{7t}{3 - 2t^2})) = -\cot(\tan^{-1}(\frac{7t}{3 - 2t^2})) = -\frac{1}{\tan(\tan^{-1}(\frac{7t}{3 - 2t^2}))} = -\frac{1}{\frac{7t}{3 - 2t^2}} = \frac{2t^2 - 3}{7t}. So t=2t2−37tt = \frac{2t^2 - 3}{7t}. 7t2=2t2−37t^2 = 2t^2 - 3 5t2=−35t^2 = -3 t2=−3/5t^2 = -3/5. This has no real solutions for tt. If k=1k=1, ϕ∈(5π4,3π2)\phi \in (\frac{5\pi}{4}, \frac{3\pi}{2}). Then 2ϕ∈(5π2,3π)2\phi \in (\frac{5\pi}{2}, 3\pi). sin⁡(2ϕ)=sin⁡(3π−2ϕ)\sin(2\phi) = \sin(3\pi - 2\phi). 3π−2ϕ∈(0,π2)3\pi - 2\phi \in (0, \frac{\pi}{2}). sin⁡−1(sin⁡(2ϕ))=3π−2ϕ\sin^{-1}(\sin(2\phi)) = 3\pi - 2\phi. 12sin⁡−1(sin⁡(2ϕ))=3π2−ϕ=3π2−tan⁡−1(t/3)\frac{1}{2}\sin^{-1}(\sin(2\phi)) = \frac{3\pi}{2} - \phi = \frac{3\pi}{2} - \tan^{-1}(t/3). Equation: θ=tan⁡−1(2t)−(3π2−tan⁡−1(t/3))=tan⁡−1(2t)+tan⁡−1(t/3)−3π2\theta = \tan^{-1}(2t) - (\frac{3\pi}{2} - \tan^{-1}(t/3)) = \tan^{-1}(2t) + \tan^{-1}(t/3) - \frac{3\pi}{2}. Since t>3t>3, tan⁡−1(2t)+tan⁡−1(t/3)=π+tan⁡−1(7t3−2t2)\tan^{-1}(2t) + \tan^{-1}(t/3) = \pi + \tan^{-1}(\frac{7t}{3 - 2t^2}). Equation: θ=π+tan⁡−1(7t3−2t2)−3π2=tan⁡−1(7t3−2t2)−3π2\theta = \pi + \tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{3\pi}{2} = \tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{3\pi}{2}. t=tan⁡(tan⁡−1(7t3−2t2)−3π2)=cot⁡(tan⁡−1(7t3−2t2))=3−2t27tt = \tan(\tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{3\pi}{2}) = \cot(\tan^{-1}(\frac{7t}{3 - 2t^2})) = \frac{3 - 2t^2}{7t}. 5t2=−35t^2 = -3, no real solutions. For other values of k≥1k \ge 1, 2ϕ2\phi will be in an interval where sin⁡−1(sin⁡(2ϕ))\sin^{-1}(\sin(2\phi)) leads to similar results.

Case 3: t<−3t < -3. If t<−3t < -3, then tan⁡ϕ<−1\tan \phi < -1, so ϕ∈(−π2+kπ,−π4+kπ)\phi \in (-\frac{\pi}{2} + k\pi, -\frac{\pi}{4} + k\pi) for integer kk. If k=0k=0, ϕ∈(−π2,−π4)\phi \in (-\frac{\pi}{2}, -\frac{\pi}{4}). Then 2ϕ∈(−π,−π2)2\phi \in (-\pi, -\frac{\pi}{2}). sin⁡(2ϕ)=sin⁡(−π−2ϕ)\sin(2\phi) = \sin(-\pi - 2\phi). −π−2ϕ∈(−π2,0)-\pi - 2\phi \in (-\frac{\pi}{2}, 0). sin⁡−1(sin⁡(2ϕ))=−π−2ϕ\sin^{-1}(\sin(2\phi)) = -\pi - 2\phi. 12sin⁡−1(sin⁡(2ϕ))=−π2−ϕ=−π2−tan⁡−1(t/3)\frac{1}{2}\sin^{-1}(\sin(2\phi)) = -\frac{\pi}{2} - \phi = -\frac{\pi}{2} - \tan^{-1}(t/3). Equation: θ=tan⁡−1(2t)−(−π2−tan⁡−1(t/3))=tan⁡−1(2t)+tan⁡−1(t/3)+π2\theta = \tan^{-1}(2t) - (-\frac{\pi}{2} - \tan^{-1}(t/3)) = \tan^{-1}(2t) + \tan^{-1}(t/3) + \frac{\pi}{2}. Since t<−3t < -3, 2t<−62t < -6 and t/3<−1t/3 < -1. x=2t<0,y=t/3<0x=2t < 0, y=t/3 < 0. xy=2t2/3>2(−3)2/3=6>1xy = 2t^2/3 > 2(-3)^2/3 = 6 > 1. Use tan⁡−1x+tan⁡−1y=−π+tan⁡−1(x+y1−xy)\tan^{-1} x + \tan^{-1} y = -\pi + \tan^{-1}(\frac{x+y}{1-xy}) for x<0,y<0,xy>1x<0, y<0, xy>1. tan⁡−1(2t)+tan⁡−1(t/3)=−π+tan⁡−1(7t3−2t2)\tan^{-1}(2t) + \tan^{-1}(t/3) = -\pi + \tan^{-1}(\frac{7t}{3 - 2t^2}). Equation: θ=−π+tan⁡−1(7t3−2t2)+π2=tan⁡−1(7t3−2t2)−π2\theta = -\pi + \tan^{-1}(\frac{7t}{3 - 2t^2}) + \frac{\pi}{2} = \tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{\pi}{2}. t=tan⁡(tan⁡−1(7t3−2t2)−π2)=−cot⁡(tan⁡−1(7t3−2t2))=2t2−37tt = \tan(\tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{\pi}{2}) = -\cot(\tan^{-1}(\frac{7t}{3 - 2t^2})) = \frac{2t^2 - 3}{7t}. 5t2=−35t^2 = -3, no real solutions. If k=1k=1, ϕ∈(π2,3π4)\phi \in (\frac{\pi}{2}, \frac{3\pi}{4}). Then 2ϕ∈(π,3π2)2\phi \in (\pi, \frac{3\pi}{2}). sin⁡(2ϕ)=sin⁡(π−2ϕ)\sin(2\phi) = \sin(\pi - 2\phi). π−2ϕ∈(−π2,0)\pi - 2\phi \in (-\frac{\pi}{2}, 0). sin⁡−1(sin⁡(2ϕ))=π−2ϕ\sin^{-1}(\sin(2\phi)) = \pi - 2\phi. 12sin⁡−1(sin⁡(2ϕ))=π2−ϕ=π2−tan⁡−1(t/3)\frac{1}{2}\sin^{-1}(\sin(2\phi)) = \frac{\pi}{2} - \phi = \frac{\pi}{2} - \tan^{-1}(t/3). Equation: θ=tan⁡−1(2t)−(π2−tan⁡−1(t/3))=tan⁡−1(2t)+tan⁡−1(t/3)−π2\theta = \tan^{-1}(2t) - (\frac{\pi}{2} - \tan^{-1}(t/3)) = \tan^{-1}(2t) + \tan^{-1}(t/3) - \frac{\pi}{2}. Since t<−3t < -3, tan⁡−1(2t)+tan⁡−1(t/3)=−π+tan⁡−1(7t3−2t2)\tan^{-1}(2t) + \tan^{-1}(t/3) = -\pi + \tan^{-1}(\frac{7t}{3 - 2t^2}). Equation: θ=−π+tan⁡−1(7t3−2t2)−π2=tan⁡−1(7t3−2t2)−3π2\theta = -\pi + \tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{\pi}{2} = \tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{3\pi}{2}. t=tan⁡(tan⁡−1(7t3−2t2)−3π2)=cot⁡(tan⁡−1(7t3−2t2))=3−2t27tt = \tan(\tan^{-1}(\frac{7t}{3 - 2t^2}) - \frac{3\pi}{2}) = \cot(\tan^{-1}(\frac{7t}{3 - 2t^2})) = \frac{3 - 2t^2}{7t}. 7t2=3−2t27t^2 = 3 - 2t^2 9t2=39t^2 = 3 t2=1/3t^2 = 1/3. t=±1/3t = \pm 1/\sqrt{3}. These values are not in the interval t<−3t < -3. For other values of k≤−1k \le -1, 2ϕ2\phi will be in an interval where sin⁡−1(sin⁡(2ϕ))\sin^{-1}(\sin(2\phi)) leads to similar results.

The only real solutions are obtained when t∈[−3,3]t \in [-3, 3], which gives t=0,1,−1t=0, 1, -1. These correspond to θ=0,π4,−π4\theta = 0, \frac{\pi}{4}, -\frac{\pi}{4}. We verified that these three values satisfy the original equation. The total number of real solutions is 3.