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Find general solution of the equation $(\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta =2 $.

My approach: Squared on both sides, formed a quadratic equation in $\cos\theta$ and finally got two solutions for theta, $$\theta = 2n\pi \pm \frac{\pi}{6}$$ $$\theta = 2n\pi \pm \frac{\pi}{3}$$ But the answer given in my book is $$2n\pi \pm \frac{\pi}{4} + \frac{\pi}{12}$$ Pretty strange
can anyone help me?

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    $\begingroup$ When you square the statement, you are introducing new roots to the original statement. It is up to you after evaluating the roots to GO BACK to the original statement and verify that they are indeed roots to the original equation, and not just the squared one. For example, if $x=2$, then I can claim $x^2=4$ then $x=2$ or $x=-2$. However, $x=-2$ doesn't satisfy my original equation. $\endgroup$ Nov 18, 2017 at 4:30

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\begin{align} (\sqrt3 - 1)\sin\theta + (\sqrt3 + 1)\cos\theta &=2\\ \left(\frac{\sqrt3}{2} -\frac{1}{2}\right)\sin\theta+\left(\frac{\sqrt3}{2} +\frac{1}{2}\right)\cos\theta&=1\\ \left(\cos30 -\sin(30)\right)\sin\theta+\left(\cos30 +\sin(30)\right)\cos\theta&=1\\ \sin\theta\cos30 -\sin\theta\sin(30)+\cos\theta\cos30 +\cos\theta\sin(30)&=1\\ \sin(\theta+30) +\cos(\theta+30) &=1\\ \sin(\theta+\pi/6) +\cos(\theta+\pi/6) &=1\\ \text{Clearly the solution set includes $\pi/3 + 2\pi k$ and $-\pi/6+2\pi k$}\\ \text{The trick now to simplify is to take the average value}\\ (\pi/3 + -\pi/6)/2&= \pi/12\\ \text{The two solutions show up every $2\pi k$}\\ \text{The distance for either solution from $\pi/12$ is $\pi/4$ }\\ \theta &= \pi/12 + 2\pi k \pm \pi/4 \end{align}

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In general, to solve $a \cos x +b\sin x=c$. Divide both sides by $\sqrt{a^2+b^2}$ and then let $\sin \alpha =\frac{a}{\sqrt{a^2+b^2}}$ to get $$\sin (x+\alpha)=\frac{c}{\sqrt{a^2+b^2}}$$ Now solve for $x$.

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$$2=(\tan60^\circ-\tan45^\circ)\sin\theta+(\tan60^\circ+\tan45^\circ)\cos\theta$$

$$\iff\dfrac1{\sqrt2}=\sin15^\circ\sin\theta+\sin105^\circ\cos\theta$$

As $\sin105^\circ=\sin(90^\circ+15^\circ)=\cos15^\circ,$

We have $\cos45^\circ=\cos(\theta-15^\circ)$

$\implies\theta-15^\circ=360^\circ n\pm45^\circ$ where $n$ is any integer

More generally, $$\sec A=(\tan A-\tan45^\circ)\sin\theta+(\tan A+\tan45^\circ)\cos\theta$$

$$\implies\cos\{\theta-(A-45^\circ)\}=\cos45^\circ$$

Here $A=60^\circ$

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