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?
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