C2 Solving Trigonometric Equations | Maths Teaching - Free Printable

To solve the given trigonometric equations for all solutions between \(0^\circ\) and \(360^\circ\), we will use the properties of the sine, cosine, and tangent functions, as well as their periodicity and symmetry. Let's solve each equation step by step.

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a) \(\sin x = 0.5\)

The sine function is positive in the first and second quadrants. The reference angle for \(\sin x = 0.5\) is:
\[
x = \arcsin(0.5) = 30^\circ
\]
Thus, the solutions are:
\[
x = 30^\circ \quad \text{and} \quad x = 180^\circ - 30^\circ = 150^\circ
\]
So, the solutions are:
\[
\boxed{30^\circ, 150^\circ}
\]

---

b) \(\cos x = 0.5\)

The cosine function is positive in the first and fourth quadrants. The reference angle for \(\cos x = 0.5\) is:
\[
x = \arccos(0.5) = 60^\circ
\]
Thus, the solutions are:
\[
x = 60^\circ \quad \text{and} \quad x = 360^\circ - 60^\circ = 300^\circ
\]
So, the solutions are:
\[
\boxed{60^\circ, 300^\circ}
\]

---

c) \(\tan x = 1\)

The tangent function is positive in the first and third quadrants. The reference angle for \(\tan x = 1\) is:
\[
x = \arctan(1) = 45^\circ
\]
Thus, the solutions are:
\[
x = 45^\circ \quad \text{and} \quad x = 180^\circ + 45^\circ = 225^\circ
\]
So, the solutions are:
\[
\boxed{45^\circ, 225^\circ}
\]

---

d) \(\sin A = -0.7\)

The sine function is negative in the third and fourth quadrants. The reference angle for \(\sin A = 0.7\) is:
\[
A = \arcsin(0.7) \approx 44.4^\circ
\]
Thus, the solutions are:
\[
A = 180^\circ + 44.4^\circ = 224.4^\circ \quad \text{and} \quad A = 360^\circ - 44.4^\circ = 315.6^\circ
\]
So, the solutions are:
\[
\boxed{224^\circ, 316^\circ}
\]

---

e) \(\cos A = -0.2\)

The cosine function is negative in the second and third quadrants. The reference angle for \(\cos A = 0.2\) is:
\[
A = \arccos(0.2) \approx 78.5^\circ
\]
Thus, the solutions are:
\[
A = 180^\circ - 78.5^\circ = 101.5^\circ \quad \text{and} \quad A = 180^\circ + 78.5^\circ = 258.5^\circ
\]
So, the solutions are:
\[
\boxed{102^\circ, 259^\circ}
\]

---

f) \(\tan A = -3\)

The tangent function is negative in the second and fourth quadrants. The reference angle for \(\tan A = 3\) is:
\[
A = \arctan(3) \approx 71.6^\circ
\]
Thus, the solutions are:
\[
A = 180^\circ - 71.6^\circ = 108.4^\circ \quad \text{and} \quad A = 360^\circ - 71.6^\circ = 288.4^\circ
\]
So, the solutions are:
\[
\boxed{108^\circ, 288^\circ}
\]

---

g) \(\sin \theta = 0.75\)

The sine function is positive in the first and second quadrants. The reference angle for \(\sin \theta = 0.75\) is:
\[
\theta = \arcsin(0.75) \approx 48.6^\circ
\]
Thus, the solutions are:
\[
\theta = 48.6^\circ \quad \text{and} \quad \theta = 180^\circ - 48.6^\circ = 131.4^\circ
\]
So, the solutions are:
\[
\boxed{48.6^\circ, 131^\circ}
\]

---

h) \(\cos \theta = -0.75\)

The cosine function is negative in the second and third quadrants. The reference angle for \(\cos \theta = 0.75\) is:
\[
\theta = \arccos(0.75) \approx 41.4^\circ
\]
Thus, the solutions are:
\[
\theta = 180^\circ - 41.4^\circ = 138.6^\circ \quad \text{and} \quad \theta = 180^\circ + 41.4^\circ = 221.4^\circ
\]
So, the solutions are:
\[
\boxed{139^\circ, 221^\circ}
\]

---

i) \(\tan \theta = 0.05\)

The tangent function is positive in the first and third quadrants. The reference angle for \(\tan \theta = 0.05\) is:
\[
\theta = \arctan(0.05) \approx 2.86^\circ
\]
Thus, the solutions are:
\[
\theta = 2.86^\circ \quad \text{and} \quad \theta = 180^\circ + 2.86^\circ = 182.86^\circ
\]
So, the solutions are:
\[
\boxed{2.86^\circ, 183^\circ}
\]

---

j) \(\tan \theta = -\frac{4}{9}\)

The tangent function is negative in the second and fourth quadrants. The reference angle for \(\tan \theta = \frac{4}{9}\) is:
\[
\theta = \arctan\left(\frac{4}{9}\right) \approx 23.8^\circ
\]
Thus, the solutions are:
\[
\theta = 180^\circ - 23.8^\circ = 156.2^\circ \quad \text{and} \quad \theta = 360^\circ - 23.8^\circ = 336.2^\circ
\]
So, the solutions are:
\[
\boxed{156^\circ, 336^\circ}
\]

---

k) \(\sin x = 0.6\)

The sine function is positive in the first and second quadrants. The reference angle for \(\sin x = 0.6\) is:
\[
x = \arcsin(0.6) \approx 36.9^\circ
\]
Thus, the solutions are:
\[
x = 36.9^\circ \quad \text{and} \quad x = 180^\circ - 36.9^\circ = 143.1^\circ
\]
So, the solutions are:
\[
\boxed{36.9^\circ, 143^\circ}
\]

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l) \(\cos x = 2\)

The cosine function has a range of \([-1, 1]\). Since \(2\) is outside this range, there are no solutions for \(\cos x = 2\).
So, the solution is:
\[
\boxed{\text{No solution}}
\]

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Final Answers:

\[
\boxed{
\begin{array}{ll}
\text{a)} & 30^\circ, 150^\circ \\
\text{b)} & 60^\circ, 300^\circ \\
\text{c)} & 45^\circ, 225^\circ \\
\text{d)} & 224^\circ, 316^\circ \\
\text{e)} & 102^\circ, 259^\circ \\
\text{f)} & 108^\circ, 288^\circ \\
\text{g)} & 48.6^\circ, 131^\circ \\
\text{h)} & 139^\circ, 221^\circ \\
\text{i)} & 2.86^\circ, 183^\circ \\
\text{j)} & 156^\circ, 336^\circ \\
\text{k)} & 36.9^\circ, 143^\circ \\
\text{l)} & \text{No solution}
\end{array}
}
\]