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I have an equation that I'm trying to solve:

$ \sin x + \sqrt 3 \cos x = 1 $

After pondering for a while and trying different things out, this chain of steps is what I ended up with:

$ \sin x + \sqrt 3 \cos x = 1 $

$ \sin x = 1 - \sqrt 3 \cos x $

$ \left(\sin x \right)^2 = \left(1- \sqrt 3 \cos x\right)^2 $

$ \sin^2 x = 1 - 2 \sqrt 3 \cos x + 3 \cos^2 x $

$ 2 \sqrt 3 \cos x - 3 \cos^2 x = 1 - \sin^2 x $

$ 2 \sqrt 3 \cos x - 3 \cos^2 x = \cos^2 x $

$ 2 \sqrt 3 \cos x = \cos^2 x + 3 \cos^2 x $

$ 4 \cos^2 x = 2 \sqrt 3 \cos x $

$ \frac{4 \cos^2 x}{\cos x} = 2 \sqrt 3 $

$ 4 \cos x = 2 \sqrt 3 $

$ \cos x = \frac{2 \sqrt 3}{4} $

$ \cos x = \frac{\sqrt 3}{2} $

The fraction $ \frac{\sqrt 3}{2} $ can be rewritten as $ \cos \left(\pm \frac{\pi}{6}\right) $, so my solutions are:

$ \cos x = \cos \left(\frac{\pi}{6}\right) \quad \text{or} \quad \cos x = \cos \left(-\frac{\pi}{6}\right) $

$ x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = -\frac{\pi}{6} + 2\pi n $

Since I earlier on exponentiated both sides I have to check my solutions:

$ x = \frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(\frac{\pi}{6} + 2\pi\right) = 2 \not = \text{RHS} $

$ x = -\frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(-\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(-\frac{\pi}{6} + 2\pi\right) = 1 = \text{RHS} $

Leaving $ x = -\frac{\pi}{6} + 2\pi n $ as the answer since its positive counterpart was not equal to $ 1 $.

$ \text{Answer:} \: x = -\frac{\pi}{6} + 2\pi n $

Have I done anything wrong or does this look good? I haven't really done this before so I feel uncertain not just about the solution, but also my steps and notation...

  • 0
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5 Answers 5

3

Note that the equation $\sin x + \sqrt 3 \cos x = 1$ is not equivalent to the equation $(\sin x)^2=(1βˆ’\sqrt{3} \cos x)^2$.

9

There is a standard method for solving equations of the form:

$ A \sin x + B \cos x = C $

Divide both sides by $\sqrt{A^2 + B^2}$:

$ \frac{A}{\sqrt{A^2 + B^2}} \sin x + \frac{B}{\sqrt{A^2 + B^2}} \cos x = \frac{C}{\sqrt{A^2 + B^2}} $

Find $\theta \in [0, 2\pi)$ so that:

$ \sin \theta = \frac{B}{\sqrt{A^2 + B^2}} \\ \cos \theta = \frac{A}{\sqrt{A^2 + B^2}} \\ $

And $\phi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ so that:

$ \sin \phi = \frac{C}{\sqrt{A^2 + B^2}} $

If you cannot find such a $\phi$, then the equation doesn't have any solutions. (For example, if $\frac{C}{\sqrt{A^2 + B^2}} > 1$.

Thus:

$ \cos \theta \sin x + \sin \theta \cos x = \sin \phi $

Using the angle sum identity, we have:

$ \sin(x + \theta) = \sin \phi $

Therefore:

\begin{align*} x_1 &= \phi - \theta + 2 \pi n \\ x_2 &= \pi - \phi - \theta + 2 \pi n \end{align*} (Where $n \in \mathbb{Z}$)


Now, let's apply this method to your question. We have:

$ A = 1, B = \sqrt{3}, C = 1 \\ \sqrt{A^2 + B^2} = 2 \\ \sin \theta = \frac{\sqrt{3}}{2}, \cos \theta = \frac{1}{2}, \theta = \frac{\pi}{3} \\ \sin \phi = \frac{1}{2}, \phi = \frac{\pi}{6} $

Thus:

$ x_1 = -\frac{\pi}{6} + 2 \pi n \\ x_2 = \frac{\pi}{2} + 2 \pi n \\ $

6

You went a little bit astray after $4 \cos^2 x = 2 \sqrt 3 \cos x$, when you divided by $\cos x$: what if $\cos x=0$?

It’s better at that point to bring everything to one side and factor: $4\cos^2x-2\sqrt3\cos x=0$, so $2\cos x(2\cos x-\sqrt3)=0$. Now appeal to the fact that if a product is $0$, at least one of the factors must be $0$. Obviously $2\ne 0$, so either $\cos x=0$, or $2\cos x-\sqrt3=0$. As it happens, both of these possibilities give you solutions. You found the second set, but not the first set.

If $\cos x=0$, we need $\sin x=1$ to have a solution. If $\sin x=1$, $\cos x$ is automatically $0$, so you just need to find the solutions to $\sin x=1$ to complete your solution.

4

There's a nice trick: $\sin x + \sqrt 3 \cos x = 1 \\\\ = 2 \left(\frac{1}{2}\sin x + \frac{\sqrt 3}{2} \cos x\right) \\\\ = 2\left(\cos\left(\frac{\pi}{3} + 2k\pi\right)\sin x + \sin\left(\frac{\pi}{3} + 2k\pi\right)\cos x \right)\\\\= 2\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1.$

When is $\sin\left(x + \frac{\pi}{3} + 2k\pi\right) = 1/2$

true?

3

You can collapse the left-hand side into a single sine function: $\sin(x)+\sqrt3\cos(x) = 2\sin(x+\pi/3)$ Then, dividing by two, all that remains is to solve the following: $\sin(x+\pi/3) = \frac{1}{2}$

Wikipedia has an article on useful trigonometric identities, including linear combinations of sin and cos.

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    Brilliant Idea !!! +1 – 2017-01-29