Solving Simple Trigonometric Equations

This page contains multiple-choice questions on solving simple trigonometric equations. Each question is followed by its correct answer and a clear explanation to help you understand the reasoning behind the solution.

Questions

Question 1

If \(0 < t < 2\pi\) such that \[ \sin t = \frac{\sqrt{2}}{2} \] and \(\cot t < 0\), find \(t\).

a) \(\pi/4\)
b) \(5\pi/4\)
c) \(7\pi/4\)
d) \(3\pi/4\)

Question 2

If \(0 < t < 2\pi\) and \[ \sin t = -1 \] find \(t\).

a) \(\pi/2\)
b) \(3\pi/2\)
c) \(5\pi/4\)
d) \(\pi\)

Question 3

If \(-2\pi < t < 0\) and \[ \sin t = -\frac{1}{2} \] find \(t\).

a) \(-5\pi/6\)
b) \(-7\pi/6\)
c) \(-5\pi/4\)
d) \(-5\pi/3\)

Question 4

If \(0 < \alpha < \pi\) and \[ \cos \alpha = -\frac{\sqrt{3}}{2} \] find \(\alpha\).

a) \(\pi/6\)
b) \(-\pi/6\)
c) \(7\pi/6\)
d) \(5\pi/6\)

Question 5

Find all values of \(t\) such that \[ \sin t = 0 \]

a) \(t = k\pi/2\)
b) \(t = k\pi/4\)
c) \(t = k\pi\)
d) \(t = 2k\pi\), where \(k \in \mathbb{Z}\)

Question 6

Find all values of \(t\) such that \[ \cos(\pi t) = 1 \]

a) \(t = 2k\pi\)
b) \(t = 2k\)
c) \(t = k\)
d) \(t = k\pi\), where \(k \in \mathbb{Z}\)

Question 7

Find all angles \(\theta\) such that \(-2\pi < \theta < 2\pi\) and \[ \cos \theta = \frac{\sqrt{2}}{2} \]

a) \(\{-7\pi/4, -\pi/4, \pi/4, 7\pi/4\}\)
b) \(\{-\pi/4, -3\pi/4, \pi/4, 3\pi/4\}\)
c) \(\{-5\pi/4, \pi/4, 3\pi/4\}\)
d) \(\{\pi/4, 3\pi/4\}\)

Answers and Explanations

1. d) \(3\pi/4\)
   The equation \(\sin t = \sqrt{2}/2\) is satisfied at \(t = \pi/4\) and \(3\pi/4\). Since \(\cot t < 0\), sine and cosine must have opposite signs, which occurs in the second quadrant. Therefore, \(t = 3\pi/4\).
2. b) \(3\pi/2\)
   The sine function reaches \(-1\) at the angle \(t = 3\pi/2\) within the interval \((0, 2\pi)\).
3. a) \(-5\pi/6\)
   The reference angle for \(\sin t = 1/2\) is \(\pi/6\). Since sine is negative and \(t\) lies between \(-2\pi\) and \(0\), the correct angle is \(-5\pi/6\).
4. d) \(5\pi/6\)
   The cosine value \(-\sqrt{3}/2\) occurs in the second quadrant with reference angle \(\pi/6\). Thus, \(\alpha = \pi - \pi/6 = 5\pi/6\).
5. c) \(t = k\pi\)
   The sine function is zero at all integer multiples of \(\pi\). Hence, the general solution is \(t = k\pi\), where \(k\) is any integer.
6. b) \(t = 2k\)
   The cosine function equals 1 when its argument is an even multiple of \(\pi\). Since \(\cos(\pi t) = 1\), we must have \(\pi t = 2k\pi\), which gives \(t = 2k\).
7. a)
   The equation \(\cos \theta = \sqrt{2}/2\) has reference angle \(\pi/4\) in quadrants I and IV. Including all such angles in \((-2\pi, 2\pi)\) gives \(-7\pi/4, -\pi/4, \pi/4, 7\pi/4\).

More Practice

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