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Question

Валентина

5 год назад

ОТВЕТЫ

Тонка

Nov 29, 2020

$sin(t)^{2} + 2sin(t) - 3 =0$

$t^{2} +2t - 3 =0$

Д = $b^{2} - 4ac = 2^{2} - 4*1*(-3) = 4 +12 = 16 = 4^{2}$

$x1 = \frac{-2 +4}{2}= 1$

$x2 = \frac{-2-4}{2}= -3$

sin(t) = -3

sin (t) = 1

$t= \frac{\pi }{2} +2k\pi$ , k⊂ Z

t = 90° + k * 360°, k⊂ Z

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$2cos(t)^{2}+ 5sin(t) -4 = 0$

$2(1-sin(t)^{2} ) + 5sin(t) - 4 = 0$

$-2-2sin(t)^{2} +5sin(t)=0$

-2-2 $t^{2} +5t=0$

Д = $5^{2} -4*(-2)*(-2) = 25 - 16 = 9 = 3^{2}$

$x1 = \frac{-5 +3}{2*(-2)}= \frac{1}{2}$

$x2 = \frac{-5-3}{2*(-2)}= 2$

sin(t) = $\frac{1}{2}$

sin(t) = 2

t = $\frac{\pi }{6} +2k\pi$ , k⊂ Z

t = $\frac{5\pi }{6} +2k\pi$ , k⊂ Z

t = $\left \{ {{30+k*360} \atop {150+k*360}} \right.$ , k⊂ Z

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