Vidaara.orgClass 11 · Mathematics
CodeVID-M11-03-TEQ-01
Trigonometric Equations — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The general solution of $\sin\theta=0$ is:
- A.$n\pi$
- B.$2n\pi$
- C.$(2n+1)\tfrac{\pi}{2}$
- D.$n\pi+\tfrac{\pi}{2}$
2.
The general solution of $\cos\theta=1$ is:
- A.$n\pi$
- B.$2n\pi$
- C.$(2n+1)\pi$
- D.$\tfrac{n\pi}{2}$
3.
If $\sin\theta=\sin\alpha$, then $\theta=$
- A.$2n\pi\pm\alpha$
- B.$n\pi+(-1)^n\alpha$
- C.$n\pi+\alpha$
- D.$2n\pi+\alpha$
4.
If $\tan\theta=\tan\alpha$, then $\theta=$
- A.$n\pi+\alpha$
- B.$2n\pi\pm\alpha$
- C.$n\pi+(-1)^n\alpha$
- D.$2n\pi+\alpha$
5.
If $\cos\theta=\cos\alpha$, then $\theta=$
- A.$n\pi+\alpha$
- B.$2n\pi\pm\alpha$
- C.$n\pi+(-1)^n\alpha$
- D.$n\pi$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
6.
Solve $\sin\theta=\tfrac12$ (general solution).
7.
Solve $\cos\theta=\tfrac12$ (general solution).
8.
Solve $\tan\theta=1$ (general solution).
9.
Solve $\sin\theta=0$ in $[0,2\pi]$.
Section C — Short Answer (3 marks)
4 × 3 = 12 marks
10.
Solve $\cos\theta=-\tfrac12$ (general solution).
11.
Solve $2\sin^2\theta-1=0$ (general solution).
12.
Solve $\sin 2\theta=\tfrac12$ (general solution).
13.
Solve $\tan\theta=\sqrt3$ in $[0,2\pi]$.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
14.
Find the general solution of $2\cos^2\theta+3\sin\theta=0$.
15.
Find the general solution of $\sqrt3\cos\theta+\sin\theta=1$.
Answer Key
Section A — Multiple Choice Questions
- (A) $n\pi$
- (B) $2n\pi$
- (B) $n\pi+(-1)^n\alpha$
- (A) $n\pi+\alpha$
- (B) $2n\pi\pm\alpha$
Section B — Short Answer (2 marks)
- $\theta=n\pi+(-1)^n\tfrac{\pi}{6}$.
- $\theta=2n\pi\pm\tfrac{\pi}{3}$.
- $\theta=n\pi+\tfrac{\pi}{4}$.
- $0,\ \pi,\ 2\pi$.
Section C — Short Answer (3 marks)
- $\theta=2n\pi\pm\tfrac{2\pi}{3}$.
- $\theta=n\pi\pm\tfrac{\pi}{4}$.
- $\theta=\tfrac{n\pi}{2}+(-1)^n\tfrac{\pi}{12}$.
- $\tfrac{\pi}{3},\ \tfrac{4\pi}{3}$.
Section D — Long Answer (5 marks)
- $\sin\theta=-\tfrac12$, so $\theta=n\pi+(-1)^n\!\left(-\tfrac{\pi}{6}\right)$.
- $\theta=2n\pi+\tfrac{\pi}{2}$ or $\theta=2n\pi-\tfrac{\pi}{6}$.
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