Mathematics, 14.07.2019 22:00 goblebrandon
Asap ! 2 sin^2 (x) -5 sin (x) -3=0 i. rewrite the equation by substituting the expression u in for sin x. ii. factor the quadratic expression. rewrite the equation with factors instead of the original polynomial. iii. use the zero product property to solve the quadratic equation. iv. rewrite your solutions to part iii by replacing u with sin x. v. solve the remaining equations for x, giving all solutions to the equation.
Answers: 1
![2\sin^2(x)-5\sin(x)-3=0\\\\Part\ I:\\u=\sin(x);\ u\in[-1;\ 1]\to2u^2-5u-3=0\\\\Part\ II:\\2u^2-6u+u-3=0\\2u(u-3)+1(u-3)=0\\(u-3)(2u+1)=0\\\\Part\ III:\\u-3=0\ \vee\ 2u+1=0\\u=3\notin[-1;\ 1]\ \vee\ 2u=-1\to u=-\dfrac{1}{2}\in[-1;\ 1]\\\\Part\ IV:\\\sin(x)=-\dfrac{1}{2}\\\\Part\ V:\\x=\dfrac{7\pi}{6}+2k\pi\ \vee\ x=-\dfrac{\pi}{6}+2k\pi\ (k\in\mathbb{Z})](/tpl/images/0090/1584/2ed15.png)
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Asap ! 2 sin^2 (x) -5 sin (x) -3=0 i. rewrite the equation by substituting the expression u in fo...
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