Question

10cos2x+3cosx=1
решите уравнение, пожалуйста.

Answer 1

$10cos(2x)+3cos(x)=1\\10cos^2(x)-10sin^2(x)+3cos(x)-sin^2(x)-cos^2(x)=0\\9cos^2(x)-11sin^2(x)+3cos(x)=0\\9cos^2(x)-11(1-cos^2(x))+3cos(x)=0\\cos(x)=t\\9t^2-11(1-t^2)+3t=0\\9t^2-11+11t^2+3t=0\\20t^2+3t-11=0\\D=880+9=889\\t_1= \frac{-3+ \sqrt{889} }{40} \\t_2= \frac{-3- \sqrt{889} }{40} \\cos(x)= \frac{-3+ \sqrt{889} }{40} \\x=+-arccos( \frac{-3+ \sqrt{889} }{40}+2\pi k \\cos(x)= \frac{-3- \sqrt{889} }{40} \\x=+-arccos( \frac{-3- \sqrt{889} }{40} )+2\pi k$
t∈[-1;1]
k∈Z