Question:

If 0º≤x<2pi, what are all the olutions of 2 sin²Θ + 3 sin Θ + 1 = 0?

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a (2/3) pi, pi/2, (1/3)pi

b (7/6) pi, (3/2)p, (11/6)pi

c (5/6) pi, pi/2, pi/6

d (4/3) pi, (3/2)pi, (5/3)pi

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  1. Treat this like a quadratic equation by making the substitution z = sin(q)

    2 sin^2q +3sin q + 1 = 0 ---&gt; 2z^2+3z+1 = 0

    Factor --&gt; (2z +1)(z+1) = 0  ---&gt; z =-1 and z = -1/2

    Then sin(q) =-1,-1/2  Now sin(q) is positive in first and second quadrants so look at angles &gt; pi.

    sin(q) = -1 when q  = 3pi/2

    sin(q) = -1/2 when q = pi+ pi/6 and q = 3pi/2+ pi/3 --&gt; q = 7pi/6, 11/6 pi

    Answer is b.


  2. Answer is b.

    Explanation:

    Let x = sin theta. Then,

    2x^2 + 3x + 1 =0

    (2x+1) (x+1)=0

    which yields

    x= -1/2 and x= -1

    Both answers are acceptable.

    For sin theta= -1/2, theta=7/6pi and 11/6pi

    For sin theta= -1, theta= 3/2pi

    Hence the answer.

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