C2 Solving Trigonometric Equations | Maths Teaching - Free Printable

We are tasked with solving the given trigonometric equations for all solutions between \(0^\circ\) and \(180^\circ\). Let's solve each equation step by step.

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a) \(\sin 2x = 0.5\)

The general solution for \(\sin \theta = 0.5\) is:
\[
\theta = 30^\circ + 360^\circ k \quad \text{or} \quad \theta = 150^\circ + 360^\circ k, \quad k \in \mathbb{Z}
\]
Here, \(\theta = 2x\), so:
\[
2x = 30^\circ + 360^\circ k \quad \text{or} \quad 2x = 150^\circ + 360^\circ k
\]
Solving for \(x\):
\[
x = 15^\circ + 180^\circ k \quad \text{or} \quad x = 75^\circ + 180^\circ k
\]
Considering \(0^\circ \leq x \leq 180^\circ\):
- For \(x = 15^\circ + 180^\circ k\): \(k = 0\) gives \(x = 15^\circ\).
- For \(x = 75^\circ + 180^\circ k\): \(k = 0\) gives \(x = 75^\circ\).

Thus, the solutions are:
\[
x = 15^\circ, 75^\circ
\]

---

b) \(\sin 3x = 1\)

The general solution for \(\sin \theta = 1\) is:
\[
\theta = 90^\circ + 360^\circ k, \quad k \in \mathbb{Z}
\]
Here, \(\theta = 3x\), so:
\[
3x = 90^\circ + 360^\circ k
\]
Solving for \(x\):
\[
x = 30^\circ + 120^\circ k
\]
Considering \(0^\circ \leq x \leq 180^\circ\):
- For \(k = 0\): \(x = 30^\circ\).
- For \(k = 1\): \(x = 150^\circ\).

Thus, the solutions are:
\[
x = 30^\circ, 150^\circ
\]

---

c) \(\cos 2x = 0.5\)

The general solution for \(\cos \theta = 0.5\) is:
\[
\theta = 60^\circ + 360^\circ k \quad \text{or} \quad \theta = 300^\circ + 360^\circ k, \quad k \in \mathbb{Z}
\]
Here, \(\theta = 2x\), so:
\[
2x = 60^\circ + 360^\circ k \quad \text{or} \quad 2x = 300^\circ + 360^\circ k
\]
Solving for \(x\):
\[
x = 30^\circ + 180^\circ k \quad \text{or} \quad x = 150^\circ + 180^\circ k
\]
Considering \(0^\circ \leq x \leq 180^\circ\):
- For \(x = 30^\circ + 180^\circ k\): \(k = 0\) gives \(x = 30^\circ\).
- For \(x = 150^\circ + 180^\circ k\): \(k = 0\) gives \(x = 150^\circ\).

Thus, the solutions are:
\[
x = 30^\circ, 150^\circ
\]

---

d) \(\cos \left(\frac{x}{2}\right) = 0.5\)

The general solution for \(\cos \theta = 0.5\) is:
\[
\theta = 60^\circ + 360^\circ k \quad \text{or} \quad \theta = 300^\circ + 360^\circ k, \quad k \in \mathbb{Z}
\]
Here, \(\theta = \frac{x}{2}\), so:
\[
\frac{x}{2} = 60^\circ + 360^\circ k \quad \text{or} \quad \frac{x}{2} = 300^\circ + 360^\circ k
\]
Solving for \(x\):
\[
x = 120^\circ + 720^\circ k \quad \text{or} \quad x = 600^\circ + 720^\circ k
\]
Considering \(0^\circ \leq x \leq 180^\circ\):
- For \(x = 120^\circ + 720^\circ k\): \(k = 0\) gives \(x = 120^\circ\).
- For \(x = 600^\circ + 720^\circ k\): No valid solutions in the range.

Thus, the solution is:
\[
x = 120^\circ
\]

---

e) \(\tan 2x = 1\)

The general solution for \(\tan \theta = 1\) is:
\[
\theta = 45^\circ + 180^\circ k, \quad k \in \mathbb{Z}
\]
Here, \(\theta = 2x\), so:
\[
2x = 45^\circ + 180^\circ k
\]
Solving for \(x\):
\[
x = 22.5^\circ + 90^\circ k
\]
Considering \(0^\circ \leq x \leq 180^\circ\):
- For \(k = 0\): \(x = 22.5^\circ\).
- For \(k = 1\): \(x = 112.5^\circ\).
- For \(k = 2\): \(x = 202.5^\circ\) (out of range).

Thus, the solutions are:
\[
x = 22.5^\circ, 112.5^\circ
\]

---

f) \(\cos 2\theta = -0.6\)

Using the inverse cosine function:
\[
2\theta = \cos^{-1}(-0.6)
\]
From a calculator:
\[
\cos^{-1}(-0.6) \approx 126.9^\circ
\]
The general solution for \(\cos \theta = -0.6\) is:
\[
2\theta = 126.9^\circ + 360^\circ k \quad \text{or} \quad 2\theta = 360^\circ - 126.9^\circ + 360^\circ k = 233.1^\circ + 360^\circ k, \quad k \in \mathbb{Z}
\]
Solving for \(\theta\):
\[
\theta = 63.45^\circ + 180^\circ k \quad \text{or} \quad \theta = 116.55^\circ + 180^\circ k
\]
Considering \(0^\circ \leq \theta \leq 180^\circ\):
- For \(\theta = 63.45^\circ + 180^\circ k\): \(k = 0\) gives \(\theta = 63.45^\circ\).
- For \(\theta = 116.55^\circ + 180^\circ k\): \(k = 0\) gives \(\theta = 116.55^\circ\).

Thus, the solutions are:
\[
\theta = 63.5^\circ, 116.6^\circ
\]

---

g) \(\tan 2\theta = -0.5\)

Using the inverse tangent function:
\[
2\theta = \tan^{-1}(-0.5)
\]
From a calculator:
\[
\tan^{-1}(-0.5) \approx -26.57^\circ
\]
Since \(\tan \theta\) is periodic with period \(180^\circ\), the general solution is:
\[
2\theta = -26.57^\circ + 180^\circ k \quad \text{or} \quad 2\theta = 180^\circ - 26.57^\circ + 180^\circ k = 153.43^\circ + 180^\circ k, \quad k \in \mathbb{Z}
\]
Solving for \(\theta\):
\[
\theta = -13.285^\circ + 90^\circ k \quad \text{or} \quad \theta = 76.715^\circ + 90^\circ k
\]
Considering \(0^\circ \leq \theta \leq 180^\circ\):
- For \(\theta = -13.285^\circ + 90^\circ k\): \(k = 1\) gives \(\theta = 76.715^\circ\).
- For \(\theta = 76.715^\circ + 90^\circ k\): \(k = 0\) gives \(\theta = 76.715^\circ\), and \(k = 1\) gives \(\theta = 166.715^\circ\).

Thus, the solutions are:
\[
\theta = 76.7^\circ, 166.7^\circ
\]

---

h) \(\tan \left(\frac{\theta}{2}\right) = \sqrt{3}\)

The general solution for \(\tan \theta = \sqrt{3}\) is:
\[
\theta = 60^\circ + 180^\circ k, \quad k \in \mathbb{Z}
\]
Here, \(\theta = \frac{\theta}{2}\), so:
\[
\frac{\theta}{2} = 60^\circ + 180^\circ k
\]
Solving for \(\theta\):
\[
\theta = 120^\circ + 360^\circ k
\]
Considering \(0^\circ \leq \theta \leq 180^\circ\):
- For \(k = 0\): \(\theta = 120^\circ\).

Thus, the solution is:
\[
\theta = 120^\circ
\]

---

i) \(\sin 4x = \frac{\sqrt{2}}{2}\)

The general solution for \(\sin \theta = \frac{\sqrt{2}}{2}\) is:
\[
\theta = 45^\circ + 360^\circ k \quad \text{or} \quad \theta = 135^\circ + 360^\circ k, \quad k \in \mathbb{Z}
\]
Here, \(\theta = 4x\), so:
\[
4x = 45^\circ + 360^\circ k \quad \text{or} \quad 4x = 135^\circ + 360^\circ k
\]
Solving for \(x\):
\[
x = 11.25^\circ + 90^\circ k \quad \text{or} \quad x = 33.75^\circ + 90^\circ k
\]
Considering \(0^\circ \leq x \leq 180^\circ\):
- For \(x = 11.25^\circ + 90^\circ k\): \(k = 0\) gives \(x = 11.25^\circ\), and \(k = 1\) gives \(x = 101.25^\circ\).
- For \(x = 33.75^\circ + 90^\circ k\): \(k = 0\) gives \(x = 33.75^\circ\), and \(k = 1\) gives \(x = 123.75^\circ\).

Thus, the solutions are:
\[
x = 11.3^\circ, 33.8^\circ, 101.3^\circ, 123.8^\circ
\]

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j) \(\tan \left(\frac{\theta}{3}\right) = 0.3\)

Using the inverse tangent function:
\[
\frac{\theta}{3} = \tan^{-1}(0.3)
\]
From a calculator:
\[
\tan^{-1}(0.3) \approx 16.70^\circ
\]
The general solution for \(\tan \theta = 0.3\) is:
\[
\frac{\theta}{3} = 16.70^\circ + 180^\circ k, \quad k \in \mathbb{Z}
\]
Solving for \(\theta\):
\[
\theta = 50.1^\circ + 540^\circ k
\]
Considering \(0^\circ \leq \theta \leq 180^\circ\):
- For \(k = 0\): \(\theta = 50.1^\circ\).

Thus, the solution is:
\[
\theta = 50.1^\circ
\]

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Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{a)} & x = 15^\circ, 75^\circ \\
\text{b)} & x = 30^\circ, 150^\circ \\
\text{c)} & x = 30^\circ, 150^\circ \\
\text{d)} & x = 120^\circ \\
\text{e)} & x = 22.5^\circ, 112.5^\circ \\
\text{f)} & \theta = 63.5^\circ, 116.6^\circ \\
\text{g)} & \theta = 76.7^\circ, 166.7^\circ \\
\text{h)} & \theta = 120^\circ \\
\text{i)} & x = 11.3^\circ, 33.8^\circ, 101.3^\circ, 123.8^\circ \\
\text{j)} & \theta = 50.1^\circ \\
\end{array}
}
\]