Solve, for $0 \leq x < 180^{\circ}$, $$ \cos(3x - 10^{\circ}) = -0.4 $$ giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 6

Using the cosine function's periodic properties, we can find other angles:

[ 3x - 10^{\circ} = 360^{\circ} - a \quad \text{or} \quad 3x - 10^{\circ} = a ]

This gives us:

[ 3x - 10^{\circ} = 360^{\circ} - 113.58^{\circ} \approx 246.42^{\circ} ]

[ 3x - 10^{\circ} = 113.58^{\circ} ]

From the first equation:

[ 3x = 246.42^{\circ} + 10^{\circ} ]

[ 3x = 256.42^{\circ} ]

[ x \approx \frac{256.42^{\circ}}{3} \approx 85.47^{\circ} \approx 85.5^{\circ} ]

From the second equation:

[ 3x = 113.58^{\circ} + 10^{\circ} ]

[ 3x = 123.58^{\circ} ]

[ x \approx \frac{123.58^{\circ}}{3} \approx 41.19^{\circ} \approx 41.2^{\circ} ]

Now consider:

[ 3x - 10^{\circ} = 360^{\circ} - 113.58^{\circ} ]

[ 3x - 10^{\circ} \approx 246.42^{\circ} ]

Solving, we find:

[ 3x = 246.42^{\circ} + 10^{\circ} \approx 256.42^{\circ} ]

[ x \approx 85.5^{\circ} ]

Lastly, from the original periodic property, we look for:

[ 3x - 10^{\circ} = 473.58^{\circ} ]

This gives:

[ 3x = 473.58^{\circ} + 10^{\circ} ]

[ 3x \approx 483.58^{\circ} ]

[ x \approx 161.19^{\circ} \approx 161.2^{\circ} ]