Hi, The question is solve for ipi =< x <= pi: $\displaystyle 2sin(x)cos(x) + sin(x) = 0$ but i get as far as $\displaystyle 2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$ and then can't get further. Am i on the right track?
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Originally Posted by aceband Hi, The question is solve for ipi =< x <= pi: $\displaystyle 2sin(x)cos(x) + sin(x) = 0$ but i get as far as $\displaystyle 2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$ and then can't get further. Am i on the right track? It may work but not the way I'd go. I'd factorise the equation $\displaystyle sin(x)(2cos(x)+1) = 0$ Either $\displaystyle sin(x)=0$ or $\displaystyle cos(x) = -\frac{1}{2}$
Originally Posted by aceband Hi, The question is solve for ipi =< x <= pi: $\displaystyle 2sin(x)cos(x) + sin(x) = 0$ but i get as far as $\displaystyle 2cos^{3}(x) + cos^{2}(x) - 2cos(x) - 1 = 0$ and then can't get further. Am i on the right track? No. You're meant to factorise: $\displaystyle \sin x (2 \cos x + 1) = 0$ Then use the null factor law to get two simple equations that are easily solved.
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