How does one solve $\sin x-\sqrt{3}\ \cos x=1$?

Hint: at the very beginning divide both sides by $2$ and use the formula for the sin of difference of 2 arguments


Hint :

$$\cos x - \sqrt{3}\sin x = 0 \Leftrightarrow \frac{\sin x}{\cos x} = \frac{\sqrt{3}}{3} \Leftrightarrow \tan x = \frac{\sqrt{3}}{3}$$

Note : You can divide by $\cos x$, since if the case was $\cos x =0$, it would be $\sin x = \pm 1$ and thus the equation would yield $\pm \sqrt{3} \neq 0$, thus no problems in the final solution, as the $\cos$ zeros are no part of it.


Multiply by the conjugate: $(\cos(x) - \sqrt{3} \sin(x))(\cos(x) + \sqrt{3} \sin(x)) = 0$. Then we have $\cos^2(x)-3\sin^2(x)=0$. This is the same thing as $1-4\sin^2(x)=0$ or $\sin(x)=\pm \frac{1}{2}$.

  • NOTE OF CAUTION: This gives you the answers to both the question and its conjugate. You'd have to plug in and check which ones are the answers you're looking for.

Tags:

Trigonometry

Algebra Precalculus

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