A)2sin^2 x - 1=0 b)6sin^2 x + sin x = 2 c)cos^2*2x - вопрос №2269787221062222132 от zhansaya12 03.12.2021 00:29

Question

Алгебра: A)2sin^2 x - 1=0 b)6sin^2 x + sin x = 2 c)cos^2*2x - 1 - cos x = корень из 3/2 - sin^2*2x

zhansaya12 · Answer

a) 2sin^2 x - 1=02sin^2 x  = 1sin^2x = 1/2sinx = ± √2/2sinx = √2/2x = pi/4 + 2pikx = 3pi/4 + 2piksinx = - √2/2x = - pi/4 +2pikx = 5pi/4 + 2pikb)6sin^2 x + sin x -  2 = 0 (3sinx + 2)( 2sinx - 1) = 0sinx = - 2/3 === x = (-1)^(k+1)*arcsin(2/3) + pik, k ∈ Zsinx = 1/2 == x = (-1)^k * pi/6 + pik, k ∈ Z...