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TRIGONOMETRY

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Can someone please show me step by step solutions? I'd appreciate if you did so I know how you get the answer. Thank you so much!

Solve each equation, giving a general formula for all of the solutions:
a)2sin^2 x-5sinx-3=0
b)(cosx-(sqrt2/2))(secx-1)=0

Would the general formula be tan x=sin x/cos x?

  • TRIGONOMETRY - ,

    2sin^2 x - 5sinx - 3 = 0
    (2sinx+1)(sinx-3) = 0
    so,
    sinx=3 or sinx = -1/2

    now, sinx is never 3, so that's out

    sin π/6 = 1/2, so that's your reference angle

    sin x < 0 in QII,QIV, so we want

    x = π + π/6
    x = 2π - π/6

    in general, then recalling that sinx has period 2π,

    x = kπ - π/6 for any integer k


    (cosx - 1/√2)(secx - 1) = 0
    so,
    cosx = 1/√2 or secx = 1

    cos π/4 = 1/√2
    sec 0 = 1

    so,

    x = 2kπ ± π/4
    x = 2kπ

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