Question

1+cos(pi/6-x)=sin^2x+0,5sinx ХЕЛП

Answer 1

Ответ:

Объяснение:

1 + cos Π/6*cos x + sin Π/6*sin x = sin^2 x + 0,5*sin x

1 + √3/2*cos x + 0,5*sin x = sin^2 x + 0,5*sin x

1 + √3/2*cos x = sin^2 x

1 - sin^2 x + √3/2*cos x = 0

cos^2 x + √3/2*cos x = 0

cos x*(cos x + √3/2) = 0

1) cos x = 0

x1 = Π/2 + Π*k; k € Z

2) cos x = -√3/2

x2 = 5Π/6 + 2Π*n; n € Z

x3 = 7Π/6 + 2Π*n; n € Z

Answer 2

Ответ:

Объяснение:

$1+cos(\frac{\pi }{6} -x) = sin^{2} x+0,5sinx;\\1+cos\frac{\pi }{6} *cosx+sin\frac{\pi }{6}*sinx =sin^{2} x +0,5sinx;\\ 1+\frac{\sqrt{3} }{2}*cosx+\frac{1}{2}*sinx-sin^{2} x -\frac{1}{2} *sinx=0;\\\\ 1+\frac{\sqrt{3} }{2} *cosx-sin^{2} x=0;\\cos ^{2}x +\frac{\sqrt{3} }{2} *cosx =0;$

$cosx( cosx+ \frac{\sqrt{3} }{2} )=0;\\\\\left[ \begin{array}{lcl} {{cosx=0} ,\\\\ {cosx= - \frac{\sqrt{3} }{2} }} \end{array} \right.$

$\left [\begin{array}{lcl} {{x=\frac{\pi }{2} +\pi n,~n\in\mathbb{Z}, } \\\\ {x=\pm\frac{5\pi }{6} }+2\pi k,~k\in\mathbb{Z}. } \end{array} \right.$