An inverse-cosine equation and factorising a trig equation

Full worked solution

1. (a) Solve ${\cos }^{-1}(2x)=\frac{2}{3}\pi$

   To solve an inverse cosine equation, we take the cosine of both sides, since $\cos \left({\cos }^{-1}(z)\right)=z$ for $-1\le z\le 1$, the valid domain of ${\cos }^{-1}$. Apply cosine to both sides:

   $$\cos\left(\cos^{-1}(2x)\right) = \cos\left(\frac{2\pi}{3}\right)$$

   We know $\cos \left(\frac{2\pi }{3}\right)=-\frac{1}{2}$, so:

   $$2x = -\frac{1}{2}$$

   Solve for $x$ by dividing both sides by 2:

   $$x = -\frac{1}{4}$$

   Verify that $2x=-\frac{1}{2}$ lies in the domain of ${\cos }^{-1}$ ($-1\le -\frac{1}{2}\le 1$), so this solution is valid. Therefore the exact solution is $x=-\frac{1}{4}$.

   Note.

   The principal value range of ${\cos }^{-1}(z)$ is $[0,\pi ]$, so we do not need to consider extra solutions here, only the principal value.

2. (b) Factorise the trigonometric equation

   Use grouping to factorise the equation $3\sin y\cos y-3\sin y-\cos y+1=0$, just as you would factorise a quadratic polynomial. Group the first two terms and the last two terms:

   $$\left(3\sin y \cos y - 3\sin y\right) + \left(-\cos y + 1\right) = 0$$

   Factor common terms from each group:

   $$3\sin y\left(\cos y - 1\right) - 1\left(\cos y - 1\right) = 0$$

   Factor out the common binomial factor $\left(\cos y-1\right)$:

   $$\left(3\sin y - 1\right)\left(\cos y - 1\right) = 0$$

   This product equals zero when either factor equals zero, so we have two separate equations to solve: $3\sin y-1=0$ and $\cos y-1=0$.

   Note.

   When factorising trigonometric expressions with two trigonometric functions, grouping almost always works if the expression is factorable.

3. (b) Solve $\cos y-1=0$ for $0\le y\le 2\pi$

   Rearrange the equation to isolate $\cos y$:

   $$\cos y = 1$$

   On the interval $[0,2\pi ]$, cosine equals 1 only at the angle where the unit circle has x-coordinate 1, which is at $y=0$ and $y=2\pi$. Therefore solutions from this factor are $y=0$ and $y=2\pi$.

   Note.

   Cosine is an even function, but we only consider angles inside the given closed interval, so we do not extend to negative angles.

4. (b) Solve $3\sin y-1=0$ for $0\le y\le 2\pi$

   Rearrange the equation to isolate $\sin y$:

   $$\sin y = \frac{1}{3}$$

   The principal solution (in $[-\frac{\pi }{2},\frac{\pi }{2}]$) is $y={\sin }^{-1}\left(\frac{1}{3}\right)$. Since sine is positive in both the first and second quadrants, the second solution in $[0,2\pi ]$ is $y=\pi -{\sin }^{-1}\left(\frac{1}{3}\right)$. Both are within the interval $[0,2\pi ]$. Therefore solutions from this factor are $y={\sin }^{-1}\left(\frac{1}{3}\right)$ and $y=\pi -{\sin }^{-1}\left(\frac{1}{3}\right)$. For decimal form, these are approximately $y\approx 0.340$ rad and $y\approx 2.802$ rad.

   Note.

   Remember that for $\sin y=k$ with $0<k<1$, there are two solutions in $[0,2\pi ]$: one acute, one obtuse.

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