# Solve for x : $\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$

Solve $$\sqrt{2}\sin(x)+\sqrt{6}\cos(x) = \sqrt{3} +1$$ for $$x$$

I started by multiplying both sides of the equation by $$\frac{1}{2\sqrt{2}}$$ to obtain

$$\displaystyle\frac{\sin(x)}{2}+\frac{\sqrt{3}\cos(x)}{2} = \frac{\sqrt{3} +1}{2\sqrt{2}}$$ $$\iff \sin(60+x) = \frac{\sqrt{3} +1}{2\sqrt{2}}$$

I am stuck here. Any hints on solving the R.H.S will be appreciated.

• You could also start by squaring both sides. You can then use double angle formulas to reduce to something that requires only the $30^\circ/60^\circ$ exact ratios. Bear in mind that squaring both sides is not a step that can be "undone" logically, so you will probably get some erroneous solutions. Just check them at the end. – Theo Bendit May 27 at 8:37

$$\sin 75^0 = \frac{\sqrt{3}+1}{2\sqrt2}$$

Can you find now?

PROOF

$$\sin (x+y) = \sin x\cos y +\cos x \sin y$$
$$\sin(75^o) = \sin(30^o+45^o) = \sin 30\cos 45 +\cos 30 \sin 45 = \frac{1}{2\sqrt2}+\frac{\sqrt3}{2\sqrt2} = \frac{\sqrt3+1}{2\sqrt2}$$

Try to find what is $$\sin\frac{5\pi}{12}$$ If you get it $$\frac{\sqrt3+1}{2\sqrt2}$$ then you did correctly and you know what to do next using the general definition of $$\sin x=\sin y$$ yields what you know it !

you can find $$\sin\frac{5\pi}{12}$$ using the formula for $$\sin \frac{x}{2}$$ by taking $$x=\frac{5\pi}{6}$$ in degrees which is equivalent to 75 degrees

• Did you arrive at $\frac{5\pi}{12}$ using a calculator? – NoLand'sMan May 27 at 8:14
• nope.. firest square it and notice what you get it after squaring .... you will get the idea. I know the formulas for sin 15 and sin 75 that's why i got it quickly.. i know half angle formulas... like for 22.5 also – learningstudent May 27 at 8:16
• if you think my solution wokrs for u .. please vote i will be obliged... – learningstudent May 27 at 8:17
• Cheers friend, an upvote from me (+1) – Ak19 May 27 at 8:32
• You're welcome🙂 – Ak19 May 27 at 8:35

square both sides, we get $$2 \sin^2(x) + 6 \cos^2(x) + 2\sqrt{12}\cos(x)\sin(x) = (1 + \sqrt{3})^2$$ that is $$2 + 4 \cos^2(x) + 2\sqrt{12}\cos(x)\sin(x) = (1 + \sqrt{3})^2$$ that is $$2 + 4 \cos^2(x) + \sqrt{12}\sin(2x) = (1 + \sqrt{3})^2$$ that is $$2 + 4 \frac{1}{1+\tan^2(x)} + \sqrt{12}\frac{2\tan x}{1+\tan^2(x)} = (1 + \sqrt{3})^2$$ Multiply everything by $$1 + \tan^2(x)$$ $$2(1+\tan^2(x)) + 4 +2 \sqrt{12}\tan x = (1 + \sqrt{3})^2(1+\tan^2(x))$$ This is quadratic in $$y = \tan(x)$$

$$ay^2 + by + c = 0$$ where \begin{align} a &= 2- (1 + \sqrt{3})^2\\ b &=2\sqrt{12}\\ c &= 2 + 4 - (1 + \sqrt{3})^2 \end{align} Solving you get something like \begin{align} y_1 &= 1.0000 \\ y_2 &= 0.2679 \end{align}

Now taking the inverse tangent of that you get \begin{align} x_1 &= \tan^{-1} y_1 =\tan^{-1} 1.0000 = 45^\circ\\ x_2 &= \tan^{-1} y_2 =\tan^{-1} 0.2679= 15^\circ \end{align}

• I think you missed multiplying 1+tanx on the R.H.S – NoLand'sMan May 27 at 8:24
• right .. ill edit – Ahmad Bazzi May 27 at 8:25

$$\frac{\sqrt{3} +1}{2\sqrt{2}}=\frac{\sqrt{3}\sqrt{2} +\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{2}}{2}\cdot\frac{1}{2}=\sin{45^\circ}\cos{30^\circ}+\cos{45^\circ}\sin{30^\circ}=\sin{(45^\circ+30^\circ)}=\sin{75^\circ}$$