How to solve this trigonometric equation $\cos(\theta +20) = -0.74$

Question

Hi my question is what approach i should use in order to solve this $\cos(\theta +20) = -0.74$?

and why do we get rid of the negative sign when solving basis angle?

Here is my working

$\cos(\theta +20) = -0.74$

Then, $\alpha+20 = \arccos (0.74)$, to find the basis angle. After I plugged into the calculator and got $42.3$, I subtract $20$ from $42.3$ to solve for $\theta$ which is $22.3$. $\theta = 42.3 -20 = 22.3$

Since $\cos$ is negative, the angle is at 2nd and 4th quadrant. So I solve for the angle $180-22.3=157.7$ and $180+22.3=202.3$

But all of this is wrong can someone explain? I think it is my approach that got me wrong, but I am unsure.

Answer 1

$$\cos(\theta+20) = -0.74$$

$$\theta+20 = \arccos (-0.74)$$

Recall the range of $\arccos x$ is $0 \leq x \leq 180$.

$$\begin{cases} Quadrant \space II \implies \theta+20 = \arccos(-0.74) \implies \theta = \arccos(-0.74)-20\\ \ \\ Quadrant \space III \implies \theta+20 = 360-\arccos(-0.74) \implies \theta = 360-\arccos(-0.74)-20 \end{cases}$$

Finally, solve for $\theta$.

For the angle in the second quadrant:

$$\theta = \arccos(-0.74)-20 \approx 137.73-20 \approx 117.73$$

For the angle in the third quadrant:

$$\theta = 360-\arccos(-0.74)-20 \approx 360-137.73-20 \approx 202.27$$

Remember that angles repeat infinitely, so each angle repeats after $360°$. If the question asks for angles in the domain $0 \leq \theta \leq 360$, your answer is complete.

If, however, you’re asked for a complete set of solutions, you have to add in $\color{blue} {360n}$ for all $\color{blue}{n \in \mathbb{Z}}$.

$$\theta \approx 117.73+360n$$

$$\theta \approx 202.27+360n$$

By changing the order and signs, you reached an incorrect answer. Try to avoid doing that and you’ll be fine with these types of questions.

Answer 2

In general we have that in degrees for $x\in[0,360°]$

$$\cos x = y \implies x=\arccos y \quad \lor \quad x=360°-\arccos y$$

therefore

$$\cos(\theta +20°) = -0.74 \implies \theta +20°=\arccos (-0.74) \quad \lor \quad \theta +20°=2\pi-\arccos (-0.74)$$

Check that the calculator is set to give the result in degree.

Answer 3

Assuming all angles are in degrees, you should avoid strange switches in the sign.

With a calculator, $\arccos(-0.74)=137.7$, so the solutions are $$ \theta+20=137.7+360k $$ or $$ \theta+20=-137.7+360k $$ The solutions in the range $[0,360]$ are therefore $137.7-20=117.7$ or $-137-20+360=202.3$.

On the other hand, if angles are measured in radians, $$ \theta+20=\arccos(-0.74)+2k\pi=2.404+2k\pi $$ or $$ \theta+20=-\arccos(-0.74)+2k\pi=-2.404+2k\pi $$ Thus $$ \theta=-20+\arccos(-0.74)+2k\pi $$ or $$ \theta=-20-\arccos(-0.74)+2k\pi $$ For the solutions in the range $[0,2\pi]$ you have to choose $k=6$ in the first case and $k=8$ in the second case, obtaining $$ 1.253\qquad\text{or}\qquad 2.729 $$