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Solve the equation on the interval [0, 2π).

cos x + 2 cos x sin x = 0
A. pi/2, 7pi/6, 3pi/2

B. pi/2, 7pi/6, 3pi/2, 11pi/6

C. 7pi/6, 11pi/6 2π

Respuesta :

LammettHash
[tex]\cos x+2\cos x\sin x=\cos x(1+2\sin x)=0[/tex]

[tex]\cos x=0[/tex] in the given interval when [tex]x=\dfrac\pi2,\dfrac{3\pi}2[/tex].

[tex]1+2\sin x=0\implies \sin x=-\dfrac12[/tex] when [tex]x=\dfrac{7\pi}6,\dfrac{11\pi}6[/tex].

Answer: -(sqrt 3/2)

Step-by-step explanation:

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