Queremos resolver:
cosa+cos(3a)=0\cos a + \cos(3a) = 0
Sabemos que:
cosx+cosy=2cos(x+y2)cos(x−y2)\cos x + \cos y = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)-
Aplicando com x=ax=a e y=3ay=3a:
cosa+cos3a=2cos(a+3a2)cos(a−3a2)\cos a + \cos 3a = 2\cos\left(\frac{a+3a}{2}\right)\cos\left(\frac{a-3a}{2}\right)=2cos(2a)cos(−a)= 2\cos(2a)\cos(-a)
Como cos(−a)=cosa\cos(-a)=\cos a:
2cos(2a)cos(a)=02\cos(2a)\cos(a)=0
cosa=0\cos a = 0
ou
cos(2a)=0\cos(2a)=0
a=π2, 3π2a=\frac{\pi}{2},\; \frac{3\pi}{2}
2a=π2+kπ2a=\frac{\pi}{2}+k\pia=π4+kπ2a=\frac{\pi}{4}+\frac{k\pi}{2}
No intervalo [0,2π][0,2\pi]:
a=π4,3π4,5π4,7π4a=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}
π4, π2, 3π4, 5π4, 3π2, 7π4\frac{\pi}{4},\; \frac{\pi}{2},\; \frac{3\pi}{4},\; \frac{5\pi}{4},\; \frac{3\pi}{2},\; \frac{7\pi}{4}
π4+π2+3π4+5π4+3π2+7π4\frac{\pi}{4}+\frac{\pi}{2}+\frac{3\pi}{4}+\frac{5\pi}{4}+\frac{3\pi}{2}+\frac{7\pi}{4}
Colocando denominador 4:
1π+2π+3π+5π+6π+7π4=24π4=6π\frac{1\pi+2\pi+3\pi+5\pi+6\pi+7\pi}{4} = \frac{24\pi}{4} =6\pi
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