Math, asked by arrrapavanipavs, 1 year ago

root3 sinx-cosx =root2

Answers

Answered by 502015736
11

sqrt3sinx - cosx = sqrt2

sqrt3/2 sinx - 1/2 cosx = sqrt2/2

sinpi/3 sinx - cospi/3 cosx = 1/sqrt2

- cos (x+ pi/3) = cospi/4

cos[ pi -x- pi/3] = cospi/4

or pi-x- pi/3 = pi/4

so x= pi-pi/3- pi/4

x = 75 deg or x= 5pi/12 answer

Answered by harendrachoubay
4

x =\dfrac{5\pi}{12} or 75°

Step-by-step explanation:

We have,

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

To find, \sqrt{3}\sin x-\cos x =\sqrt{2}=?

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

Dividing both sides by 2, we get

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

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

\sin \dfrac{\pi}{3} \sin x-\cos \dfrac{\pi}{3} \cos x =\cos \dfrac{\pi}{4}

-\cos (x+ \dfrac{\pi}{3}) =\cos \dfrac{\pi}{4}

\pi -x-\dfrac{\pi}{3}=\dfrac{\pi}{4} or

x=\pi-\dfrac{\pi}{3}-\dfrac{\pi}{4}

x =\dfrac{5\pi}{12} or 75°

Hence, x =\dfrac{5\pi}{12} or 75°

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