Найдите sin a, если cos a = -корень15/4 и 90 градусов < a < 180 градусов

[tex]cos\alpha=-\frac{\sqrt{15}}{4}\\sin\alpha> 0\\sin\alpha=\sqrt{1-cos^2\alpha}\\sin\alpha=\sqrt{{\sqrt{15}}{4})^2}\\sin\alpha=\sqrt{1-\frac{15}{16}}\\sin\alpha=\sqrt{\frac{1}{16}}\\sin\alpha=\frac{1}{4}[/tex]

x^2+4x+1=(x^2+4x+4)-4+1=(x+2)^2 - 3

q(x)=2x-5

q(x+3)=2x-5+3=2x-2

q(x+3)=3x

3x=2x-2 => x=-2