Differentiation (Part-38) | S N De

Problem 1(xxxi)

$$ \frac{\sin\alpha}{\sin2x}+\frac{\cos\alpha}{\cos2x}=2 \\ \Rightarrow \frac{\sin\alpha\cos2x+\cos\alpha\sin2x}{\sin2x\cos2x}=2 \\ \Rightarrow \sin(\alpha+2x)=2\sin2x\cos2x \\ \Rightarrow \sin(\alpha+2x)=\sin4x \\ \Rightarrow \sin4x-\sin(\alpha+2x)=0 \\ \Rightarrow 2\cos\left(3x+\frac{\alpha}{2}\right)\sin\left(x-\frac{\alpha}{2}\right)=0 \\ \therefore \cos\left(3x+\frac{\alpha}{2}\right)=0 \quad \text{--- (1)}, \\ \text{or,} \quad \sin\left(x-\frac{\alpha}{2}\right)=0 \quad \text{--- (2)} $$

From (1), we get:

$$ 3x+\frac{\alpha}{2}=(2n+1)\frac{\pi}{2} \\ \Rightarrow x=(2n+1)\frac{\pi}{6}-\frac{\alpha}{6}, \quad n\in\mathbb{Z} \quad \text{--- (3)} $$

Again, by (2), we get:

$$ x-\frac{\alpha}{2}=n\pi \\ \Rightarrow x=n\pi+\frac{\alpha}{2}, \quad n\in\mathbb{Z} \quad \text{--- (4)} $$

Hence, from (3) and (4), we get the required general solution.

Problem 1(xxxii)

$$ \sqrt{3}\tan\theta\tan\left(\theta+\frac{\pi}{3}\right) \times \tan\left(\theta+\frac{2\pi}{3}\right)=1 \\ \Rightarrow \tan\theta \cdot \frac{\tan\theta+\sqrt{3}}{1-\sqrt{3}\tan\theta} \cdot \frac{\tan\theta-\sqrt{3}}{1+\sqrt{3}\tan\theta}=\frac{1}{\sqrt{3}} \\ \Rightarrow \tan\theta \cdot \frac{\tan^2\theta-3}{1-3\tan^2\theta}=\frac{1}{\sqrt{3}} \\ \Rightarrow \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}=-\frac{1}{\sqrt{3}} \\ \Rightarrow \tan 3\theta=\tan\left(-\frac{\pi}{6}\right) \\ \Rightarrow 3\theta=n\pi-\frac{\pi}{6} \\ \Rightarrow \theta=\frac{n\pi}{3}-\frac{\pi}{18}, \quad n\in\mathbb{Z}. $$

Problem 1(xxxiii)

$$ (2+\sqrt{3})\cos\theta=1-\sin\theta \\ \Rightarrow \sin\theta+(2+\sqrt{3})\cos\theta=1 \\ \Rightarrow \frac{1}{2\sqrt{2+\sqrt{3}}} \sin\theta+\frac{2+\sqrt{3}}{2\sqrt{2+\sqrt{3}}}\cos\theta=\frac{1}{2\sqrt{2+\sqrt{3}}} \\ \Rightarrow \cos(\theta-\alpha)=\sin\alpha $$

Let, $\displaystyle \frac{1}{2\sqrt{2+\sqrt{3}}}=\sin\alpha$

$$ \Rightarrow \cos(\theta-\alpha)=\cos\left(\frac{\pi}{2}-\alpha\right) \\ \Rightarrow \theta-\alpha=2n\pi \pm \left(\frac{\pi}{2}-\alpha\right) \\ \therefore \theta=2n\pi+\frac{\pi}{2}, \quad 2n\pi-\frac{\pi}{2}+2\alpha $$

where $\displaystyle \frac{1}{2\sqrt{2+\sqrt{3}}}=\sin\alpha, \quad n \in\mathbb{Z}$.

Problem 1(xxxiv)

$$ \cos2\theta=(\sqrt{2}+1)\left(\cos\theta-\frac{1}{\sqrt{2}}\right) \\ \Rightarrow 2\cos^2\theta-1=(\sqrt{2}+1)\cos\theta-1-\frac{1}{\sqrt{2}} \\ \Rightarrow 2\cos^2\theta-(\sqrt{2}+1)\cos\theta+\frac{1}{\sqrt{2}}=0 \\ \Rightarrow 2\sqrt{2}\cos^2\theta-(2+\sqrt{2})\cos\theta+1=0 \\ \Rightarrow 2\cos\theta(\sqrt{2}\cos\theta-1)-1(\sqrt{2}\cos\theta-1)=0 \\ \Rightarrow (\sqrt{2}\cos\theta-1)(2\cos\theta-1)=0 \\ \therefore \sqrt{2}\cos\theta-1=0 \quad \text{--- (1)}, \\ \text{or,} \quad 2\cos\theta-1=0 \quad \text{--- (2)} $$

From (1), we get:

$$ \cos\theta=\frac{1}{\sqrt{2}} =\cos\left(\pm \frac{\pi}{4}\right) \\ \Rightarrow \theta=2n\pi \pm \frac{\pi}{4}, \quad n\in\mathbb{Z} \quad \text{--- (3)} $$

By (2), we get:

$$ \cos\theta=\frac{1}{2}=\cos\left(\pm \frac{\pi}{3}\right) \\ \Rightarrow \theta=2n\pi \pm \frac{\pi}{3}, \quad n\in\mathbb{Z} \quad \text{--- (4)} $$

Hence, from (3) and (4), we get the required general solution.

Problem 1(xxxv)

Solve: $\cos3\theta+\cos2\theta =\sin(3\theta/2)+\sin(\theta/2) \quad \text{where} \quad 0 \leq \theta \leq 2\pi$

$$ \Rightarrow 2\cos\left(\frac{3\theta+2\theta}{2}\right) \cos\left(\frac{3\theta-2\theta}{2}\right)=2\sin\theta\cos\frac{\theta}{2} \\ \Rightarrow 2\cos\left(\frac{5\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)=2\sin\theta\cos\left(\frac{\theta}{2}\right) \\ \Rightarrow 2\cos\left(\frac{\theta}{2}\right)\left[\cos\frac{5\theta}{2}-\sin\theta\right]=0 \\ \Rightarrow \cos\left(\frac{\theta}{2}\right)=0 \quad \text{--- (1)}, \\ \text{or,} \quad \cos\left(\frac{5\theta}{2}\right)-\sin\theta=0 \quad \text{--- (2)} $$

From (1), we get:

$$ \frac{\theta}{2}=(2n+1)\frac{\pi}{2} \Rightarrow \theta=(2n+1)\pi \quad \text{--- (3)} $$

Again, by (2), we get:

$$ \cos\left(\frac{5\theta}{2}\right)=\cos\left(\frac{\pi}{2}-\theta\right) \\ \Rightarrow \frac{5\theta}{2}=2n\pi \pm \left(\frac{\pi}{2}-\theta\right) \\ \therefore \theta=\frac{(4n+1)\pi}{7} \quad \text{--- (4)}, \quad \text{and} \quad \theta=\frac{(4n-1)\pi}{3} \quad \text{--- (5)} $$

By (3) we get: $\theta=\pi$ for $(n=0)$, and $\theta=3\pi, -\pi$ for $(n=\pm 1)$

By (4) we get: $\theta=\pi/7$ $(n=0)$, $\theta=5\pi/7, -3\pi/7$ $(n=\pm1)$, $\theta= 9\pi/7, -\pi$ $(n=\pm 2)$, $\theta=13\pi/7, -11\pi/7$ $(n=\pm3)$, $\theta=17\pi/7, -15\pi/7$ $(n=\pm4)$

Finally, by (5) we get: $\theta=-\pi/3$ $(n=0)$, $\theta=\pi, -5\pi/3$ $(n=\pm1)$

$$ \therefore \theta=\pi, \frac{\pi}{7}, \frac{5\pi}{7}, \frac{9\pi}{7}, \frac{13\pi}{7} \quad [0 \leq \theta \leq 2\pi] $$

Problem 1(xxxvi)

$$ \cos3x+\sin\left(2x-\frac{7\pi}{6}\right)=-2 $$

Since $-1\leq \sin\theta,\cos\theta \leq 1$, this is only possible if:

$$ \cos3x=-1 \quad \text{--- (1)}, \quad \text{and} \quad \sin\left(2x-\frac{7\pi}{6}\right)=-1 \quad \text{--- (2)} $$

From (1), we get:

$$ 3x=(2n+1)\pi \Rightarrow x=\frac{2n\pi}{3}+\frac{\pi}{3}, \quad n\in\mathbb{Z} $$

Again, by (2), we get:

$$ 2x-\frac{7\pi}{6}=\frac{(4n-1)\pi}{2} \Rightarrow x=n\pi+\frac{\pi}{3}, \quad n\in\mathbb{Z} $$

Hence, both (1) and (2) will satisfy if:

$$ x=2n\pi+\frac{\pi}{3}, \quad n\in\mathbb{Z} $$

Problem 1(xxxvii)

$$ \sin x+\cos2x=-2 $$

Since $-1 \leq \sin \theta,\cos \theta \leq 1$, this requires:

$$ \sin x=-1 \quad \text{--- (1)}, \quad \text{and} \quad \cos2x=-1 \quad \text{--- (2)} $$

Now, from (1), we get:

$$ x=\frac{(4n-1)\pi}{2}, \quad n\in\mathbb{Z} \quad \text{--- (3)} $$

Again, by (2), we get:

$$ 2x=(2n+1)\pi \Rightarrow x=\frac{(2n+1)\pi}{2}, \quad n\in\mathbb{Z} \quad \text{--- (4)} $$

Hence, from (3) and (4), we get the required general solution.

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