Saturday

August 23, 2014

August 23, 2014

Posted by **Kate** on Saturday, May 18, 2013 at 10:31am.

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 -
**Steve**, Saturday, May 18, 2013 at 11:36am2sin^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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