Respuesta :
Answer:
The factored form is (sin x +2)(sin x-1)
Step-by-step explanation:
We have been given the trigonometric function [tex]\sin^2 x +\sinx-2[/tex]
We can factor this by AC method. In AC method we multiply the term a and c and then write the middle term b in such a way that the sum/difference is equal to the product 'ac'
Using the method, we can write sinx as 2sinx -sinx
[tex]\sin^2 x +2\sinx-\sin x-2[/tex]
Now, we group the first two terms and the last two terms
[tex](\sin^2 x +2\sinx)+(-\sin x-2)[/tex]
Now, we take GCF from each group
[tex]\sin x(\sin x +2)-1(\sin x+2)[/tex]
Factor out (sinx+2)
[tex](\sin x +2)(\sin x-1)[/tex]
Therefore, the factored form is (sin x +2)(sin x-1)
Answer:
The factor form is [tex](\sin x+2)(\sin x-1)[/tex]
Step-by-step explanation:
Given : Algebraic expression [tex]\sin^2x+\sin x-2[/tex]
To find : Factor the algebraic expression in terms of a single trigonometric function ?
Solution :
Algebraic expression [tex]\sin^2x+\sin x-2[/tex]
Let [tex]\sin x=y[/tex]
So, [tex]y^2+y-2[/tex]
To factor the expression we equate it to zero,
[tex]y^2+y-2=0[/tex]
Apply middle term split,
[tex]y^2+2y-y-2=0[/tex]
[tex]y(y+2)-1(y+2)=0[/tex]
[tex](y+2)(y-1)=0[/tex]
Substitute the value of y, [tex]y=\sin x[/tex]
[tex](\sin x+2)(\sin x-1)=0[/tex]
The factor form is [tex](\sin x+2)(\sin x-1)[/tex]
