looks like A to me
Then everything is in terms of cosx
The equation 2sinx+sqrt(3)cotx=sinx is partially solved below.
2sinx+sqrt(3)cotx=sinx
sinx(2sinx+sqrt(3)cotx)=sinx(sinx)
2sin^2x+sqrt(3)cosx=sin^2x
sin^2x+sqrt(3)cosx=0
Which of the following steps could be included in the completed solution?
a. cos^2x-sqrt(3)cosx-1=0
b. sin^2x-sqrt(3)sinx-1=0
c. (1-cos^2x)(sqrt(3)cosx)=0
d. (1-sin^2x)(sqrt(3)sinx)=0
7 answers
No Steve is actually correct, A, or cos^2x - sqrt3 cos x - 1 = 0 is correct
1. A [ cos^2 (x) - sqrt(3)*cos(x) - 1 = 0 ]
2. B [ f(x) = sqrt(sin x) ]
3. D [ 7.6 hours ]
100% ur welcome
2. B [ f(x) = sqrt(sin x) ]
3. D [ 7.6 hours ]
100% ur welcome
Tysm Yuh!
You're amazing @Yuh !!! Thank you so much<3 100%
Trigonometric Equations Practice:
1. (A): cos^2 (x) - sqrt(3)*cos(x) - 1 = 0
2. (B): f(x) = sqrt(sin x)
3. (D): 7.6 hours
Trigonometric Equations:
1. (C): three
2. (B): The equation was factored incorrectly.
3. (A & E) sec x= -1 & sec x=3/2
4. (B): sqrt2/2
1. (A): cos^2 (x) - sqrt(3)*cos(x) - 1 = 0
2. (B): f(x) = sqrt(sin x)
3. (D): 7.6 hours
Trigonometric Equations:
1. (C): three
2. (B): The equation was factored incorrectly.
3. (A & E) sec x= -1 & sec x=3/2
4. (B): sqrt2/2
^^ 100% correct! :)
1 answer