➡ find the number of solution of sin²x - sinx - 1 =0 in [-2π,2π].
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Answered by
0
Answer:
let sin x = k
k^2-k-1=0
k^2-k+k-1=0. (factorization)
k(k-1)+1(k-1)=0
(k-1)(k+1)=0
k+1=0
k=-1. (not possible)
k-1=0
k=1
sin x=1
sin x= sin 90
x=90
Answered by
30
let sin x = k
k^2-k-1=0
k^2-k+k-1=0. (factorization)
k(k-1)+1(k-1)=0
(k-1)(k+1)=0
k+1=0
k=-1. (not possible)
k-1=0
k=1
sin x=1
sin x= sin 90
x=90
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