could you help me please... √3.cosX – sinX = √3 (Worried)

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- May 6th 2009, 12:59 AMnezima√3.cosX – sinX = √3, How To Solve This?
**could you help me please... √3.cosX – sinX = √3 (Worried)**

- May 6th 2009, 01:58 AMbebrave
firstly divide all sgrt3

then when you see sgrt 3 you will be write tanpi/3 then you can solve...or if you want i can solve all - May 6th 2009, 02:03 AMnezima
can you please do it ? (Itwasntme) (Blush)

- May 6th 2009, 02:08 AMbebrave
ok..wait me pls 10 minute..;)

- May 6th 2009, 02:15 AMbebrave
i have like this question and solution you can check attachment then you can solve your problem

- May 6th 2009, 11:20 AMMoo
Hi,

There's a simpler way.

When you don't know where to start, when given $\displaystyle a\cos(x)+b\sin(x)$, find the modulus, that is $\displaystyle \sqrt{a^2+b^2}$ and factor it out.

You would then have $\displaystyle \sqrt{a^2+b^2} \left(\frac{a}{\sqrt{a^2+b^2}} \cos(x)+\frac{b}{\sqrt{a^2+b^2}} \sin(x)\right)$

Define $\displaystyle c=\frac{a}{\sqrt{a^2+b^2}}$ and $\displaystyle d=\frac{b}{\sqrt{a^2+b^2}}$

they're both in $\displaystyle [-1,1]$ and are such that $\displaystyle c^2+d^2=1$

Thus there exists $\displaystyle \theta \in [0,2\pi[$ such that $\displaystyle c=\cos \theta$ and $\displaystyle d=\sin \theta$

Hence $\displaystyle a\cos x+b\sin x=\cos\theta\cos x+\sin\theta\sin x=\cos(x-\theta)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Here, you can find that $\displaystyle \sqrt{a^2+b^2}=2$

So you can divide by 2 on both sides :

$\displaystyle \frac{\sqrt{3}}{2}\cos(x)-\frac 12 \sin(x)=\frac{\sqrt{3}}{2}$

Note that $\displaystyle \cos\left(-\tfrac\pi 6\right)=\tfrac{\sqrt{3}}{2}$ and $\displaystyle \sin\left(-\tfrac\pi 6\right)=-\tfrac 12$

So your equation is now $\displaystyle \cos\left(-\tfrac\pi 6\right)\cos(x)+\sin\left(-\tfrac\pi 6\right)\sin(x)=\tfrac{\sqrt{3}}{2}$

That is $\displaystyle \cos\left(x+\tfrac\pi 6\right)=\tfrac{\sqrt{3}}{2}$

Then remember that $\displaystyle \cos(a)=\cos(b) \Leftrightarrow \forall k\in \mathbb{Z} ~,~ a=\pm b+2k\pi$