Re-learning basics of trigonometric functions to solve $2\cos(x+10º) = 1$

Question

I'm helping my daughter with her math homework. It's been 25 years since I learned this myself. What do I need to re-learn about trigonometric functions to solve: $$2\cos(x+10º)=1$$

Answer 1

$$2\cos(x+10º) = 1$$ $$\cos(x+10º) = \frac 12$$ You have to remember that $\cos(60)=\cos(-60)=1/2$ $$\cos(x+10º) = \cos(60)$$ $$x+10=60$$ $$x=50°$$

And $$\cos(x+10º) = \cos(-60)$$ $$x+10=-60$$ $$x=-70°$$ $$x=360-70=290°$$

These are the solutions to the equation in the interval $[0,360]$ as Taussig has pointed out...

Answer 2

Write your equation as $$\cos(x+10^{\circ})=\frac{1}{2}$$ substitute $$t=x+10^{\circ}$$ and you have to solve $$\cos(t)=\frac{1}{2}$$

Answer 3

You'd need to relearn the trig tables and associated inverse trig tables.

To solve your question ;

$2\cos(x+10^{\circ}) = 1$

$\cos(x+10^{\circ})=\frac12$

$x+10^\circ = \arccos(\frac12)$

$x= 60^\circ-10^\circ $

$x = 50^\circ$ $\quad$ if $x\in \{0,\pi\} $

more generally,

$x=2n\pi\pm50^\circ$ $\quad$ for any $n\in Z$

Answer 4

\begin{align} 2\cos(x+10) &= 1 \qquad &\text{given}\\ \cos(x+10) &= 1/2 \qquad &\text{try to isolate the trig function}\\ x+10 &= \arccos(1/2) \qquad &\text{use the inverse trig function}\\ x+10 &=60, -60 \qquad &\text{consider unit circle to check for multiple values}\\ x&= 50, -70 (\text{ or $290$)} \qquad &\text{solve for $x$} \end{align}

Note that $\cos(t) = 1/2$ has two solution sets: $60 + 2\pi \cdot k$ and $-60 + 2\pi \cdot k$ where $k$ is an integer value. This is because $\cos(t)$ is periodic. I have only included the solutions from the domain $[0, 360]$

Answer 5

You should use the general solution.

You have $$\cos(x+10)=\frac 12$$ $$\implies x+10=\pm60+n\cdot360$$ $$\implies x=50\;\text{or}-70+n\cdot360$$