IIT JEE 1982 Maths - 'Adapted' - Subjective to Multi-Correct Q4

Since the problem asks about the coordinates and number of intersection points, the objective is to determine the solutions for the following $\cos(x) = \sin(3x)$

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Trigonometry Arithmetic

Applying the following triple-angle identity $\sin(3x) = 3\sin(x) - 4\sin^3(x)$ transforms the given equation to $\cos(x) = 3\sin(x) - 4\sin^3(x)$ Squaring both sides, $\begin{array}{rlccl} \left(\cos(x)\right) &= \left(3\sin(x) - 4\sin^3(x)\right)^2\\ \cos^2(x) &= 9\sin^2(x) - 24\sin^4(x) + 16\sin^6(x)\\ 1 - \sin^2(x) &= 9\sin^2(x) - 24\sin^4(x) + 16\sin^6(x) & & & {\color{#3D99F6}\text{Pythagorean identity:}\sin^2(x) + \cos^2(x) = 1}\\ 0 &= 16\sin^6(x) - 24\sin^4(x) + 10\sin^2(x) - 1 \end{array}$ Letting $u = \sin^2(x)$ , we have $0 = 16u^3 - 24u^2 + 10u - 1$ Factorizing, $\begin{array}{rl} 0 &= 16u^3 + \left(-8u^2 - 16u^2\right) + \left(8u + 2u\right) - 1\\ &= \left(16u^3 - 8u^2\right) + \left(-16u^2 + 8u\right) + \left(2u - 1\right)\\ &= 8u^2\left(2u - 1\right) - 8u\left(2u - 1\right) + 1\left(2u - 1\right)\\ &= \left(2u - 1\right)\left(8u^2 - 8u + 1\right) \end{array}$ Either (1) $2u - 1 = 0$ or (2) $8u^2 - 8u + 1 = 0$ .

Case 1

If $2u - 1 = 0$ , then $\begin{array}{rl} u &= \dfrac{1}{2}\\ \sin^2(x) &= \dfrac{1}{2}\\ \sin(x) &= \pm \dfrac{1}{\sqrt{2}}\\ \end{array}$ If $\sin(x) = -\dfrac{1}{\sqrt{2}}$ , then $x = -\dfrac{\pi}{4}$ . But since $\cos > 0$ and $\sin < 0$ at that point, this solution does not work.

Otherwise, if $\sin(x) = \dfrac{1}{\sqrt{2}}$ , then $x = \dfrac{\pi}{4}$ .

Case 2

Otherwise, $8u^2 - 8u + 1 = 0$ . Completing the squares , $\begin{array}{rlccl} 8\left(u^2 - u + \dfrac{1}{8}\right) &= 0\\ u^2 - u + \dfrac{1}{4} + \dfrac{1}{8} - \dfrac{1}{4} &= 0 & & & {\color{#3D99F6}\text{Constant factor excluded}}\\ \left(u - \dfrac{1}{2}\right)^2 &= \dfrac{1}{8} & & & {\color{#3D99F6}\text{Factor }u^2 - u + \dfrac{1}{4}\text{ and arrange terms}}\\ u &= \dfrac{1}{2} \pm \dfrac{\sqrt{2}}{4} & & & {\color{#3D99F6}\text{Result after few steps}}\\ &= \dfrac{2 \pm \sqrt{2}}{4} \end{array}$ Instead of setting both sides by $\arcsin$ , consider the following half-angle trigonometry identity: $\sin\left(\dfrac{\alpha}{2}\right) = \pm \sqrt{\dfrac{1 - \cos(\alpha)}{2}}$ Squaring both sides, $\sin^2\left(\dfrac{\alpha}{2}\right) = \dfrac{1}{2}\left(1 - \cos(\alpha)\right)$ With this type of relationship, we can express $\sin^2(x)$ as $\sin^2(x) = \dfrac{1}{2}\left(1 \pm \dfrac{\sqrt{2}}{2}\right) = \dfrac{1}{2}\left(1 - \cos(2x)\right)$ where $\begin{array}{rlccl} 1 \pm \dfrac{\sqrt{2}}{2} &= 1 - \cos(2x)\\ \cos(2x) &= \pm \dfrac{\sqrt{2}}{2}\\ 2x &= \pm\dfrac{\pi}{4} + \pi k & & & {\color{#3D99F6}\text{For }k \in \mathbb{Z}}\\ x &= \pm\dfrac{\pi}{8} + \dfrac{k\pi}{2}\\ x &= \left\{\dfrac{\pi}{8}, -\dfrac{3\pi}{8}\right\} & & & {\color{#3D99F6}\text{Finding all solutions for }-\dfrac{\pi}{2} \leq x \leq \dfrac{\pi}{2}} \end{array}$

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Results

There are 3 intersection points: $\begin{array}{rl} \text{abscissa}_1 = \dfrac{\pi}{4},&\, \text{ordinate}_1 = \dfrac{1}{\sqrt{2}}\\ \text{abscissa}_2 = \dfrac{\pi}{8},&\, \text{ordinate}_2 = \dfrac{\sqrt{2 + \sqrt{2}}}{2}\\ \text{abscissa}_3 = \dfrac{-3\pi}{8},&\, \text{ordinate}_3 = \dfrac{\sqrt{2 - \sqrt{2}}}{2} \end{array}$ so this satisfies $\boxed{\text{Statement A}}$ .

Because there are two positive abscissae and one negative, Statement B does not hold.

Since $\left(\text{ordinate}_2\right)^2 + \left(\text{ordinate}_3\right)^2 = 1$ the set of solutions also satisfies $\boxed{\text{Statement C}}$ .

Clearly, since the sum of abscissas is $0$ , $\boxed{\text{Statement D}}$ holds.