Co-aontaran triantanachd le foirmle tuinn

Eisimpleir

Fuasgail \sqrt 3 \cos x + \sin x = \sqrt 2, airson 0 \le x \le 2\pi.

Fuasgladh

An toiseach feumaidh sinn \sqrt 3 \cos x + \sin x a chur dhan riochd k\cos (x - \alpha).

k\cos (x-\alpha) = k\cos x\cos \alpha + k\sin x\sin \alpha

k\cos \alpha  = \sqrt 3

k\sin \alpha  = 1

Obraich a-mach k a' cleachdadh luachan airson nan co-èifeachdan:

k = \sqrt {\left( {\sqrt {{3^2}} } \right) + {1^2}}

= \sqrt {3 + 1}

= \sqrt 4

= 2

Bhon a tha \tan \theta  = \frac{{sin\theta }}{{\cos \theta }}

\tan \alpha  = \frac{1}{{\sqrt 3 }}

\alpha  = \frac{\pi }{6}

Mar sin \sqrt 3 \cos x + \sin x = 2\cos \left( {x - \frac{\pi }{6}} \right)

2\cos \left( {x - \frac{\pi }{6}} \right) = \sqrt 2

\cos \left( {x - \frac{\pi }{6}} \right) = \frac{{\sqrt 2 }}{2}

\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{{\sqrt 2 }}

Bhon a tha cos dearbhte, tha sinn sa 1d agus sa 4mh cairteal.

A' chiad chairteal

x - \frac{\pi }{6} = {\cos ^{ - 1}}\frac{1}{{\sqrt 2 }}

x - \frac{\pi }{6} = \frac{\pi }{4}

x = \frac{{5\pi }}{{12}}

An ceathramh cairteal

x - \frac{\pi }{6} = 2\pi  - \frac{\pi }{4}

x - \frac{\pi }{6} = \frac{{7\pi }}{4}

x = \frac{{23\pi }}{{12}}

Mar sin x = \frac{{5\pi }}{{12}},\,\frac{{23\pi }}{{12}}

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