hi there its me agfain MRMATHHEWMATICS!!!
i am sorry but i meant to write for question1 that 2cos2x is is g(x)
and that y=-squareroot 3 is the f(x)
q1) 2cos2x is f(x)
-square root3 is g(x)
solve f(x)-g(x)=0 to get points of intersection in x is between 0 and 180degrees
q2) a) write the equation cos2x + 8cosx+9=0 in terms of cosx and show that for cosx it has equal roots
q2b) show that there are no real roots for x.
for q2 i have tried to do it but i get upto the bit 2(cosx+2)(2cosx+2)=0 and i don't know what to do next.
could you please do all of question 1 as i don't get it at all and i feel if i see all then working then only then will i understand the method
2 answers
#1
So you want to solve f(x)-g(x)=0 ?
-squareroot 3 - 2cos2x = 0
cos2x = -√3/2 , I just rearranged.
The cosine is negative in II and III, so
2x = 150º or 2x = 210º
x = 75º or x = 105º for 0 ≤ x ≤ 180º
So you want to solve f(x)-g(x)=0 ?
-squareroot 3 - 2cos2x = 0
cos2x = -√3/2 , I just rearranged.
The cosine is negative in II and III, so
2x = 150º or 2x = 210º
x = 75º or x = 105º for 0 ≤ x ≤ 180º