Solve the trigonometric equation below

[chwala]

Homework Statement

Solve the equation, $$cos ∅+ \sqrt 3⋅ sin ∅=1$$ in the interval, $$0≤∅≤2π$$

Relevant Equations

understanding of equations of the form $$a sin b+c cos d=e$$

Solve the equation, $$cos ∅+ \sqrt 3⋅ sin ∅=1$$ in the interval, $$0≤∅≤2π$$

 

[chwala]

Find the question and its solution, below...this is pretty clear and easy for me...the only thing to note is that for cosine, $$cos ∅=cos (-∅)$$

ok, i also noted that we could use the half angle property here for $$tan∅$$, using this property it follows that,

$$1-t^2+ \sqrt 12⋅ t=1+t^2$$
→$$-2t^2+\sqrt 12⋅ t=0$$
$$t=1.73$$ and $$t=0$$
Therefore, $$tan \frac {1}{2}∅=1.732050808$$
$$∅={0, 120^0, 300^0}$$ which can be re-written in rad.

Any other approach guys...or we only have this two?...cheers

 

[DaveE]

OK, but I'm not sure what your question is.

I hate trig identities since I only remember a couple of them. In my experience this sort of problem is either impossible (i.e. numeric solutions only) or there are multiple ways of solving it.

Anyway, I used $cos(\theta)^2 + sin(\theta)^2=1$ and then arranged things into the quadratic $2cos(\theta)^2 - cos(\theta) -1 = 0$ which is easy to solve for $cos(\theta)$. Maybe not as nice as the solution in the book, but it's what I can do easily from memory.

I think it's odd that the domain is $0 \leq \theta \leq 2\pi$. Everyone else would choose $0 \leq \theta \lt 2\pi$. I guess they want to make sure your paying attention to the details.

 

[chwala]

OK, but I'm not sure what your question is.

I hate trig identities since I only remember a couple of them. In my experience this sort of problem is either impossible (i.e. numeric solutions only) or there are multiple ways of solving it.

Anyway, I used $cos(\theta)^2 + sin(\theta)^2=1$ and then arranged things into the quadratic $2cos(\theta)^2 - cos(\theta) -1 = 0$ which is easy to solve for $cos(\theta)$. Maybe not as nice as the solution in the book, but it's what I can do easily from memory.

I think it's odd that the domain is $0 \leq \theta \leq 2\pi$. Everyone else would choose $0 \leq \theta \lt 2\pi$. I guess they want to make sure your paying attention to the details.

Thanks Dave, it wasn't a question rather just asking if there are other methods in solving this kind of trig. equations...cheers mate.

 

[chwala]

$$cos^2 ∅+ sin^2 ∅=1$$, would work...one would have to re-express either $$cos∅$$ in terms of $$sin∅$$ or conversely, ...then express it in the form of $$ax^2+bx=1$$, then square both sides and work to solution.