CBSE Class 11 Mathematics Trigonometric Functions Worksheet Set 07

CBSE Class 11 Mathematics Worksheet - Trigonometric Functions (7). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. Solve the equation or find the general solutions \( \cos x + \cos(2x) + \cos(3x) = 0 \)
Answer: We have, \( [\cos(3x) + \cos x] + \cos(2x) = 0 \)
\( \implies 2\cos(2x) \cdot \cos x + \cos(2x) = 0 \)
\( \implies \cos(2x) [2\cos x + 1] = 0 \)
\( \implies \cos(2x) = 0 \)
\( \implies 2x = (2n + 1)\frac{\pi}{2} \)
\( \implies x = (2n + 1)\frac{\pi}{4} \)
\( \implies n \in \mathbb{Z} \)
Or,
\( 2\cos x + 1 = 0 \)
\( \implies \cos x = -\frac{1}{2} \)
\( \implies \cos x = \cos\left(\pi - \frac{\pi}{3}\right) \)
\( \implies \cos x = \cos\left(\frac{2\pi}{3}\right) \)
Comparing with \( \cos\theta = \cos\alpha \) here, \( \theta = x \) and \( \alpha = \frac{2\pi}{3} \)
\( \theta = 2n\pi \pm \alpha \)
\( \implies x = 2n\pi \pm \frac{2\pi}{3} \)
\( \therefore x = (2n + 1)\frac{\pi}{4} \) and \( x = 2n\pi \pm \frac{2\pi}{3} \); \( n \in \mathbb{Z} \) ans.

Question. Solve, \( \sin(2x) + \cos(x) = 0 \).
Answer: We have, \( \sin(2x) + \cos(x) = 0 \)
\( \implies 2\sin x \cdot \cos x + \cos x = 0 \)
\( \implies \cos x [2\sin x + 1] = 0 \)
\( \implies \cos x = 0 \)
\( \implies x = (2n + 1)\frac{\pi}{2} \)
Or,
\( 2\sin x + 1 = 0 \)
\( \implies \sin x = -\frac{1}{2} \)
\( \implies \sin x = \sin\left(\pi + \frac{\pi}{6}\right) \)
\( \implies \sin x = \sin\left(\frac{7\pi}{6}\right) \)
Comparing with \( \sin\theta = \sin\alpha \) here, \( \theta = x \) and \( \alpha = \frac{7\pi}{6} \)
\( \theta = n\pi + (-1)^n \alpha \)
\( \implies x = n\pi + (-1)^n \frac{7\pi}{6} \)
\( \therefore x = (2n + 1)\frac{\pi}{2} \) and \( x = n\pi + (-1)^n \frac{7\pi}{6} \); \( n \in \mathbb{Z} \) ans.

Question. Solve the equation: \( \sin x - 3\sin(2x) + \sin(3x) = \cos x - 3\cos(2x) + \cos(3x) \)
Answer: We have, \( \sin x - 3\sin(2x) + \sin(3x) = \cos x - 3\cos(2x) + \cos(3x) \)
\( \implies (\sin(3x) + \sin x) - 3\sin(2x) = (\cos(3x) + \cos x) - 3\cos(2x) \)
\( \implies 2\sin(2x) \cdot \cos x - 3\sin(2x) = 2\cos(2x)\cos x - 3\cos(2x) \)
\( \implies \sin(2x)(2\cos x - 3) = \cos(2x)(2\cos x - 3) \)
\( \implies \sin(2x)(2\cos x - 3) - \cos(2x)(2\cos x - 3) = 0 \)
\( \implies (2\cos x - 3)(\sin(2x) - \cos(2x)) = 0 \)
Either \( 2\cos x - 3 = 0 \) OR \( \sin(2x) - \cos(2x) = 0 \)
For \( 2\cos x - 3 = 0 \):
\( \implies \cos x = \frac{3}{2} \) (not possible since \( -1 \le \cos\theta \le 1 \))
For \( \sin(2x) - \cos(2x) = 0 \):
\( \implies \sin(2x) = \cos(2x) \)
\( \implies \tan(2x) = 1 \)
\( \implies \tan(2x) = \tan\left(\frac{\pi}{4}\right) \)
Here, \( \theta = 2x \); \( \alpha = \frac{\pi}{4} \)
\( \implies \theta = n\pi + \alpha \)
\( \implies 2x = n\pi + \frac{\pi}{4} \)
\( \implies x = \frac{n\pi}{2} + \frac{\pi}{8} \); \( n \in \mathbb{Z} \) ans.

Question. Solve, \( 2\cos^2 x + 3\sin x = 0 \).
Answer: We have, \( 2\cos^2 x + 3\sin x = 0 \)
\( \implies 2(1 - \sin^2 x) + 3\sin x = 0 \)
\( \implies 2 - 2\sin^2 x + 3\sin x = 0 \)
\( \implies 2\sin^2 x - 3\sin x - 2 = 0 \)
\( \implies 2\sin^2 x - 4\sin x + \sin x - 2 = 0 \)
\( \implies 2\sin x(\sin x - 2) + 1(\sin x - 2) = 0 \)
\( \implies (2\sin x + 1)(\sin x - 2) = 0 \)
Either \( 2\sin x + 1 = 0 \) OR \( \sin x - 2 = 0 \)
For \( \sin x - 2 = 0 \):
\( \implies \sin x = 2 \) (not possible since \( -1 \le \sin\theta \le 1 \))
For \( 2\sin x + 1 = 0 \):
\( \implies \sin x = -\frac{1}{2} \)
\( \implies \sin x = \sin\left(\pi + \frac{\pi}{6}\right) \)
\( \implies \sin x = \sin\left(\frac{7\pi}{6}\right) \)
Here, \( \theta = x \); \( \alpha = \frac{7\pi}{6} \)
\( \implies \theta = n\pi + (-1)^n \alpha \)
\( \implies x = n\pi + (-1)^n \frac{7\pi}{6} \); \( n \in \mathbb{Z} \) ans.

Question. Solve, \( \cot^2 \theta + \frac{3}{\sin\theta} + 3 = 0 \).
Answer: We have, \( \cot^2 \theta + \frac{3}{\sin\theta} + 3 = 0 \)
\( \implies (\csc^2 \theta - 1) + 3\csc \theta + 3 = 0 \)
\( \implies \csc^2 \theta + 3\csc \theta + 2 = 0 \)
\( \implies \csc^2 \theta + 2\csc \theta + \csc \theta + 2 = 0 \)
\( \implies \csc \theta(\csc \theta + 2) + 1(\csc \theta + 2) = 0 \)
\( \implies (\csc \theta + 1)(\csc \theta + 2) = 0 \)
Either \( \csc \theta = -1 \) OR \( \csc \theta = -2 \)
If \( \csc \theta = -1 \):
\( \implies \sin \theta = -1 \)
\( \implies \theta = n\pi + (-1)^n \frac{3\pi}{2} \)
If \( \csc \theta = -2 \):
\( \implies \sin \theta = -\frac{1}{2} \)
\( \implies \theta = n\pi + (-1)^n \frac{7\pi}{6} \); \( n \in \mathbb{Z} \) ans.

Question. Solve the equation: \( \tan(2x) = -\cot\left(x + \frac{\pi}{3}\right) \)
Answer: We have, \( \tan(2x) = -\cot\left(x + \frac{\pi}{3}\right) \)
\( \implies \tan(2x) = \tan\left(\frac{\pi}{2} + \left(x + \frac{\pi}{3}\right)\right) \)
\( \implies \tan(2x) = \tan\left(\frac{5\pi}{6} + x\right) \)
Here \( \theta = 2x \); \( \alpha = \frac{5\pi}{6} + x \)
\( \theta = n\pi + \alpha \)
\( \implies 2x = n\pi + \frac{5\pi}{6} + x \)
\( \implies 2x - x = n\pi + \frac{5\pi}{6} \)
\( \implies x = n\pi + \frac{5\pi}{6} \); \( n \in \mathbb{Z} \) ans.

Question. Solve, \( \sin(3x) + \cos(2x) = 0 \).
Answer: We have, \( \sin(3x) + \cos(2x) = 0 \)
\( \implies \cos(2x) = -\sin(3x) \)
\( \implies \cos(2x) = \cos\left(\frac{\pi}{2} + 3x\right) \)
Comparing with \( \cos\theta = \cos\alpha \)
\( \implies \theta = 2x \); \( \alpha = \frac{\pi}{2} + 3x \)
\( \implies \theta = 2n\pi \pm \alpha \)
\( \implies 2x = 2n\pi \pm \left(\frac{\pi}{2} + 3x\right) \)

Case 1:
\( 2x = 2n\pi + \frac{\pi}{2} + 3x \)
\( \implies -x = 2n\pi + \frac{\pi}{2} \)
\( \implies x = -\left(2n\pi + \frac{\pi}{2}\right) \); \( n \in \mathbb{Z} \) ans.

Case 2:
\( 2x = 2n\pi - \frac{\pi}{2} - 3x \)
\( \implies 5x = 2n\pi - \frac{\pi}{2} \)
\( \implies x = \frac{2n\pi}{5} - \frac{\pi}{10} \); \( n \in \mathbb{Z} \) ans.

Question. Solve, \( \sqrt{3}\cos x + \sin x = \sqrt{2} \).
Answer: We have, \( \sqrt{3}\cos x + \sin x = \sqrt{2} \)
Here, \( a = \sqrt{3} \) and \( b = 1 \)
Divide both sides by \( \sqrt{3 + 1} = 2 \)
\( \implies \frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x = \frac{\sqrt{2}}{2} \)
\( \implies \cos\left(\frac{\pi}{6}\right)\cos x + \sin\left(\frac{\pi}{6}\right)\sin x = \cos\left(\frac{\pi}{4}\right) \)
\( \implies \cos\left(x - \frac{\pi}{6}\right) = \cos\left(\frac{\pi}{4}\right) \)      ............ \( \{ \cos A \cos B + \sin A \sin B = \cos(A - B) \} \)
Form: \( \cos\theta = \cos\alpha \)
\( \theta = x - \frac{\pi}{6} \); \( \alpha = \frac{\pi}{4} \)
\( \theta = 2n\pi \pm \alpha \)
\( \implies x - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{4} \)
\( \implies x = 2n\pi \pm \frac{\pi}{4} + \frac{\pi}{6} \)

Case 1:
\( x = 2n\pi + \frac{\pi}{4} + \frac{\pi}{6} \)
\( \implies x = 2n\pi + \frac{5\pi}{12} \); \( n \in \mathbb{Z} \) ans.

Case 2:
\( x = 2n\pi - \frac{\pi}{4} + \frac{\pi}{6} \)
\( \implies x = 2n\pi - \frac{\pi}{12} \); \( n \in \mathbb{Z} \) ans.

Question. Solve the equation, \( \cot\theta + \csc\theta = \sqrt{3} \).
Answer: We have, \( \cot\theta + \csc\theta = \sqrt{3} \)
\( \implies \frac{\cos\theta}{\sin\theta} + \frac{1}{\sin\theta} = \sqrt{3} \)
\( \implies \cos\theta + 1 = \sqrt{3}\sin\theta \)
\( \implies \cos\theta - \sqrt{3}\sin\theta = -1 \)
Here, \( a = 1 \) and \( b = -\sqrt{3} \)
Divide both sides by \( \sqrt{a^2 + b^2} = \sqrt{1 + 3} = 2 \)
\( \implies \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta = -\frac{1}{2} \)
\( \implies \cos\left(\frac{\pi}{3}\right)\cos\theta - \sin\left(\frac{\pi}{3}\right)\sin\theta = \cos\left(\pi - \frac{\pi}{3}\right) \)
\( \implies \cos\left(\theta + \frac{\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) \)
Here, \( \alpha = \frac{2\pi}{3} \)
\( \implies \theta + \frac{\pi}{3} = 2n\pi \pm \frac{2\pi}{3} \)
\( \implies \theta = 2n\pi \pm \frac{2\pi}{3} - \frac{\pi}{3} \)

Case 1:
\( \theta = 2n\pi + \frac{2\pi}{3} - \frac{\pi}{3} \)
\( \implies \theta = 2n\pi + \frac{\pi}{3} \)

Case 2:
\( \theta = 2n\pi - \frac{2\pi}{3} - \frac{\pi}{3} \)
\( \implies \theta = 2n\pi - \pi \); \( n \in \mathbb{Z} \) ans.

Question. Solve, \( \tan\theta + \tan(2\theta) + \tan(3\theta) = \tan\theta \tan(2\theta) \tan(3\theta) \).
Answer: We have, \( \tan\theta + \tan(2\theta) + \tan(3\theta) = \tan\theta \tan(2\theta) \tan(3\theta) \)
\( \implies \tan\theta + \tan(2\theta) = -\tan(3\theta) + \tan\theta \tan(2\theta) \tan(3\theta) \)
\( \implies \tan\theta + \tan(2\theta) = -\tan(3\theta)[1 - \tan\theta \tan(2\theta)] \)
\( \implies \frac{\tan\theta + \tan(2\theta)}{1 - \tan\theta \tan(2\theta)} = -\tan(3\theta) \)
\( \implies \tan(\theta + 2\theta) = -\tan(3\theta) \)
\( \implies \tan(3\theta) + \tan(3\theta) = 0 \)
\( \implies 2\tan(3\theta) = 0 \)
\( \implies \tan(3\theta) = 0 \)
\( \implies 3\theta = n\pi \)
\( \implies \theta = \frac{n\pi}{3} \); \( n \in \mathbb{Z} \) ans.