SOLUTION: Find, to the nearest degree, the positive acute angle or angles that satisfy the equation: 5sin^2x +1 = 6sin x 3cos^2x +2 = 7 cos x 3 sin^2x +4sinx = 4 sin A = 1-2sin^A

Question 919539: Find, to the nearest degree, the positive acute angle or angles that satisfy the equation:
5sin^2x +1 = 6sin x
3cos^2x +2 = 7 cos x
3 sin^2x +4sinx = 4
sin A = 1-2sin^A
7sinx = 4+2sin^x
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find, to the nearest degree, the positive acute angle or angles that satisfy the equation:
5sin^2x +1 = 6sin x
5sin^2(x) - 6sin(x) + 1 = 0
Factor::
(5sin(x)-1)(sin(x)-1) = 0
sin(x) = 1/5 or sin(x) = 1
x = arcsin(1/5) or x = arcsin(1)
x = 11.54 degrees or x = 90 degrees
---------------------------------------
3cos^2x +2 = 7 cos x
3cos^2(x) - 7cos(x) + 2 = 0
(3cos(x)-1)(cos(x)-2) = 0
cos(x) = 1/3 or cos(x) = 2
arccos(1/3) = 70.53 degrees or x-value is extraneous
------------------------
3 sin^2x +4sinx = 4
3sin^2(x) + 4sin(x) - 4 = 0
(3sin(x)-2)(sin(x)+2) = 0
sin(x) = 2/3 or sin(x) = -2
x = arcsin(2/3) = 41.81 degrees or x-value is extraneous
====================================
sin A = 1-2sin^A
2sin^2(A) - sin(A) - 1 = 0
Factor::
(2sin(A) +1)(sin(A)-1) = 0
sin(A) = -1/2 or sin(A) = 1
A = arcsin(-1/2) = -30 degrees or A = 90 degrees
-----------------------------------
7sinx = 4+2sin^x
2sin^2(x) - 7sin(x) -4 = 0
(2sin(x)+1)(sin(x)-4) = 0
sin(x) = -1/2 or sin(x) = 4
x = arcsin(-1/2) = -30 degrees or x value is extraneous
=================
Cheers,
Stan H.
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