solve a trigonometric equation $\sqrt{3} \sin(x)-\cos(x)=\sqrt{2}$

Question

$$\sqrt{3}\sin{x} - \cos{x} = \sqrt{2} $$ I think to do : $$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{2}}$$ but i dont get anything. Or to divied by $\sqrt{3}$ : $$\frac{(\sqrt{3}\sin{x} - \cos{x} = \sqrt{2})}{\sqrt{3}}$$

Answer 1

Hint:

$$\sqrt3\sin x-\cos x=\sqrt2\iff \sin\frac\pi3\sin x-\cos\frac\pi3\cos x=\frac{\sqrt2}2$$

Answer 2

One of the R-Formulas, a set of formulas for combining such trigonometric expressions, says that

$$a\sin{x} - b\cos{x} = R\sin(x - \alpha)$$

where

$$R = \sqrt{a^2 + b^2}, \alpha = \tan^{-1}{\frac{b}{a}}$$

However, I doubt this solution expresses any room for creativity for this question specifically, as noted by Don Antonio's interesting observation.

Answer 3

A difficulty of this equation is that it contains both $\sin(x)$ and $\cos(x)$. You can use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ to express $\sin(x) = \pm \sqrt{1 - \cos^2(x)}$ (or $\cos(x)$ as a function of $\sin(x)$), but it is very unwieldy because it introduces square roots (another difficulty) and you need to distinguish the intervals where $\sin(x)$ is positive or negative. It is simpler if you can recognize another trigonometric identity.

The equation looks a bit like $\cos(u) \sin(x) - \sin(u) \cos(x) = a$ for some values $u$ and $a$ to be determined, only with a multiplicative factor. If you had this, you could apply the sine-difference formula: the equation is equivalent to $\sin(x-u) = a$. Your idea to multiply by a constant was on the right; the other part of the puzzle is this identity which guides you towards a multiplicative constant that helps. The equation constrains $\dfrac{\cos(u)}{\sin(u)} = \dfrac{\sqrt 3}{1}$ which you should recognize as having the solution $u = \frac{\pi}{6}$. Since $\cos(\frac{\pi}{6}) = \frac{\sqrt 3}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$, multiply the original equation by $\frac{1}{2}$ to get $$ \sin\left(x - \frac{\pi}{6}\right) = \frac{\sqrt 2}{2} $$

Alternatively, to find the multiplicative coefficient, you can remark that $\cos^2(u) + \sin^2(u) = 1$, while here you have $(\sqrt 3)^2 + (1)^2 = 4$. Thus you need to divide the equation by $\sqrt 4$ to get coefficients that are a (cos, sin) pair. (Here I am essentially deriving the formula shown by Yiyuan Lee from a more common identity.)

Since $\frac{\sqrt 2}{2} = \sin(\frac{\pi}{4})$, the equation is equivalent to $$ x - \frac{\pi}{6} = \frac{\pi}{4} + 2k\pi \qquad\text{or}\qquad x - \frac{\pi}{6} = \pi - \frac{\pi}{4} + 2k\pi $$ i.e. $x = \frac{5\pi}{12} + 2k\pi$ or $x = \frac{11\pi}{12} + 2k\pi$.

Answer 4

$$\sqrt{3} \sin x - \cos x = \sqrt{2}$$

Dividing both sides by 2, we get

$$\frac{\sqrt{3}}{2}\sin x - \frac{1}{2}\cos x = \frac{\sqrt{2}}{2}$$

By substituting $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ & $\sin 30^{\circ} = \frac{1}{2}$, we get

$$\sin x \cos30^{\circ} - \cos x \sin 30^{\circ} = \frac{\sqrt{2}}{2}$$

Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and $\sin 45^{\circ} = \frac{\sqrt{2}}{2}$,

$$\begin{align*}\sin(x - 30^{\circ}) &= \sin 45^{\circ} \\ x - 30^{\circ} &= 45^{\circ} \\ x &= 75^{\circ}\end{align*}.$$

Also, we know that $\sin 135^{\circ} = \frac{\sqrt{2}}{2}$. Then, $$\begin{align*}\sin(x - 30^{\circ}) &= \sin 135^{\circ} \\ x - 30^{\circ} &= 135^{\circ} \\ x &= 165^{\circ}\end{align*}.$$

Therefore, $x = 30^{\circ}$ and $x = 135^{\circ}$.

Answer 5

HINT:

Weierstrass substitution is not a bad alternative either.

It will leave a Quadratic Equation in $\displaystyle\tan\frac x2$ on substitution and rearrangement of the given relation.