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)

 3 cos 2x + sin x= 1 
→3 cos 2x + sin x− 1=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°

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pi/3 letter b

did it on e2020