Solve: 2 cos^2 x = 13 sin x-5?

Answer 1

Hi,

Since sin² + cos² = 1, then cos² x = 1 - sin² x, so:

2 cos² x = 13 sin x-5
2 (1 - sin² x) = 13 sin x - 5
2 - 2sin² x = 13 sin x - 5
0 = 2sin² x + 13 sin x - 7
0 = (2sin x - 1)(sin x + 7)
Setting each factor equal to zero and solving gives:

2sin x - 1 = 0
2 sin x = 1
sin x = ½
arcsin ½ or sin^-1(½) = 30° ± 360k and 150° ± 360k for all integers k

sin x + 7 = 0
sin x = -7
Since sine values only range from -1 to 1, this answer makes no sense and can be discarded.

So x = 30° ± 360k and 150° ± 360k for all integers k

I hope that helps!! :-)

Answer 2

2 (1 - sin ² x) = 13 sin x - 5
2 - 2 sin ² x = 13 sin x - 5
2 sin ² x + 13 sin x - 7 = 0
(2 sin x - 1) (sin x + 7) = 0
sin x = 1 / 2 is acceptable answer
x = 30 ° , x = 150 °

Answer 3

Put sin^2 x = 1 - sin^2 x

So the equation becomes :
2 - 2sin^2 x = 13sin x - 5
or 2sin^2 x + 13sin x -7 =0

Solve for sin x by quadratic formula.
Find x.

Hope this helps.

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Answer 4

2(1-sin^2x)=13*sinx - 5
2*sin^2x+13*sinx-7=0
2sinx(sinx+7) -1 (sinx+7)=0
(2sinx-1)(sinx+7)=0
x=sin^-1(1/2) or x=sin^-1(7)

Answer 5

Thank you