<span>2sin(π + x) – 5cos(</span>π/2 + x)<span><span> + 2 = 0</span></span><span><span />
2 sin</span>²<span>x</span><span> – 5sinx
+ 2 = 0</span>
Вводим промежуточную переменную
t = sinx
<span>2t</span>²<span> -
5t + 2 = 0
</span><span> а = 2; b = -5; c = 2<span>
</span>D = b² - 4ac = (-5)² - 4 *2 * 2 = 25 - 16 = 9<span>
</span>t1 = <u>-</u><u> b</u><u /><u> </u><u>+</u><u> </u><u>√</u><u>D</u><u /><u> </u> = <u>-</u><u> </u><u>( -
5) +</u><u> </u><u>√9 </u><u> </u> = <u /><u> </u><u>5 +</u><u> </u><u>3 </u><u> </u> =
2 НЕ принадлежит интервалу [-1; 1]</span><span><span>
2</span>a 2 * 2 4<span>
</span></span><span>t2 = <u>-</u><u> b</u><u /><u> </u><u>+</u><u> </u><u>√</u><u>D</u><u /><u> </u> = <u>-</u><u> </u><u>( - 5)
-</u><u> </u><u>√9 </u><u> </u> = <u /><u> </u><u>5</u><u> </u><u>- </u><u> </u><u>3 </u><u> </u><span> = <u> 1 </u></span></span><span>
2a 2 * 1 <span>4 2</span></span>
<span>sinx
= 1/2
<span /></span><span>x = (-1)^k (</span>π/6)<span><span> + πk,
kͼZ</span><span>
</span></span>
в сичлителе 3а- за скобку= 3а(а-2)
2х-2+(15/2-х)=28
2х-2+7.5-х=28
х=28+2-7.5
х=22.5=22.5