A trig equation with CAST (multiple angle)

Trigonometry

A-level Maths (9MA0)

ORIGINAL

Change the interval to one for 2x first, use CAST to find every angle in it, and only halve at the end.

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Question

Solve

5\cos 2x + 3 = 0

for

0 \le x < 360^\circ

, giving your answers to 1 decimal place.

Every step worked, with the reasoning.

1
$\cos 2x = -\dfrac{3}{5} = -0.6$
Make $\cos 2x$ the subject.
2
$0 \le x < 360^\circ$ , so $0 \le 2x < 720^\circ$
Double the interval: solving for $2x$ means finding every value of $2x$ in two full turns.
3
Reference angle: $\alpha = \cos^{-1}(0.6) = 53.130\ldots^\circ$
Use the positive value to get the acute angle; the sign decides the quadrants.
4
$\cos$ is negative in the second and third quadrants, so $2x = 180^\circ - \alpha = 126.869\ldots^\circ$ or $2x = 180^\circ + \alpha = 233.130\ldots^\circ$
CAST: cosine is positive only in the first (All) and fourth (Cos) quadrants.
5
Add $360^\circ$ : $2x = 486.869\ldots^\circ$ or $2x = 593.130\ldots^\circ$
Both are still below $720^\circ$ , so they are in the interval too.
6
$x = 63.434\ldots^\circ,\ 116.565\ldots^\circ,\ 243.434\ldots^\circ,\ 296.565\ldots^\circ$
Halve every value of $2x$ , and round only at the end.

Answer: $x = 63.4^\circ,\ 116.6^\circ,\ 243.4^\circ,\ 296.6^\circ$ (1 d.p.)