Solving Trigonometric Equations for the ESAT

Solving trigonometric equations requires a methodical approach to ensure every valid solution is found within the specified interval. Whether using a graphical method or a CAST diagram, the key is to account for the periodic nature of the functions: sine and cosine repeat every $360^{\circ}$ (or $2\pi$ radians), while tangent repeats every $180^{\circ}$ (or $\pi$ radians).

Solving Equations with Transformed Arguments

When an equation involves a transformed argument, such as $\sin(2x + 60)$, it is essential to find all possible values for the entire bracket before performing any algebraic rearrangement to find $x$. This prevents the common error of losing solutions.

Worked Example: Square Roots and Expanded Intervals

Consider the equation $\sin^2(2x + 60) = \frac{1}{4}$ for the range $-360 < x < 360$.

First, we take the square root of both sides. It is critical to remember that this produces both positive and negative results. This leads to two separate equations:

1. $\sin(2x + 60) = \frac{1}{2}$
2. $\sin(2x + 60) = -\frac{1}{2}$

For the first equation, the basic solution (or principal value) is found by taking $\sin^{-1} \frac{1}{2}$, which is $30^{\circ}$. By looking at the sine graph or a CAST diagram, we find the next solution is $180 - 30 = 150^{\circ}$.

To find the full set of solutions for $x$ in the range $-360$ to $360$, we must look at an expanded range for the bracket $(2x + 60)$. We list several solutions for the bracket before rearranging:

$2x + 60 = -690, -570, -330, -210, 30, 150, 390, 510, 750, 870$

Now we subtract 60 from each and divide by 2 to find $x$:

$x = -375, -315, -195, -135, -15, 45, 165, 225, 345, 405$

Finally, we filter for values that fall within the original interval $-360 < x < 360$:

$x = -315, -195, -135, -15, 45, 165, 225, 345$

The Danger of Early Rearrangement

A frequent mistake is to solve for $x$ immediately after finding just one basic solution. For instance, if you take $2x + 60 = 30$ to get $x = -15$ and then try to find general solutions for $x$ by adding multiples of $180$, you will miss half of the valid solutions because the symmetry of the sine function was not fully applied to the bracket first.

As shown in the graph above, this incorrect method fails to capture the full set of intersections between the sine curve and the constant line.

Using Trigonometric Identities

In more complex problems, an equation may contain both sine and cosine terms or appear in a quadratic form. In these cases, you must use identities to homogenise the equation into a single trigonometric function. The most common identity used for this purpose is $\sin^2 x + \cos^2 x = 1$.

Worked Example: Quadratic Trigonometric Equations

Solve $12 \cos^2 x + 6 \sin x - 10 = 2$ for the interval $0^{\circ} < x < 360^{\circ}$.

This equation looks like a quadratic but has two different functions. We use $\cos^2 x = 1 - \sin^2 x$ to transform it into an equation involving only sine:

$12(1 - \sin^2 x) + 6 \sin x - 10 = 2$

Expanding the brackets gives $12 - 12\sin^2 x + 6 \sin x - 10 = 2$. Rearranging to standard quadratic form:

$12\sin^2 x - 6 \sin x = 0$

We can let $S = \sin x$ to make the factorisation clearer:

$6S(2S - 1) = 0$

This gives two paths:

1. $S = 0$, so $\sin x = 0$. In the range $0 < x < 360$, this gives $x = 180^{\circ}$.
2. $2S - 1 = 0$, so $\sin x = \frac{1}{2}$. This gives $x = 30^{\circ}$ and $x = 180 - 30 = 150^{\circ}$.

The full solution set is $x = 30^{\circ}, 150^{\circ}, 180^{\circ}$.