Consider the equation 3 cos 2x + sin x = 1
(a) Write this equation in the form f(x) = 0, where f(x) = a sin^2 x + b sin x + c
(b) factor f(x)
(c) Write down the number of solutions of f(x) = 0, for 0 < x <2π (do not solve 0)
i.e 3 {1−2sin²x} + sin x− 1=0 → 3 −6sin²x + sin x− 1=0 → −6sin²x + sin x+2=0 →f(x) = −6sin²x + sin x+2=0 where p= −6, q=1 and r=2
f(x) = −6sin²x + sin x+2= −6sin²x +4 sin x−3sinx+2 =−2sinx(3sinx −2)−(3sinx−2) =(3sinx −2)(−2sinx−1) =−(3sinx −2)(2sinx+1)
f(x) = −6sin²x + sin x+2=0 =−(3sinx −2)(2sinx+1)=0
Either (3sinx −2) =0 i. sinx =2/3 or (2sinx+1)=0 →sinx=−1/2 0 ≤ x ≤ 2π there are two solutions for sinx =2/3
x= 41.81° and 138.19° and another two solutions for sinx=− 1/2
x= 210° 330°
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
