SOLUTION: Good morning: I am trying to help my son with the following problem as he has an exam later today and we have been working in solving the following problem: Solve algebraical

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Question 932052: Good morning:
I am trying to help my son with the following problem as he has an exam later today and we have been working in solving the following problem:
Solve algebraically for θ in the first quadrant: (cosθ)^2 = sinθ -.02
Thank you.
Caroline
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Use a substitution,
cos%5E2%28theta%29=1-sin%5E2%28theta%29.
Then,
1-sin%5E2%28theta%29=sin%28theta%29-.02
sin%5E2%28theta%29%2Bsin%28theta%29-1.02=0
Now use another substitution,
u=sin%28theta%29
u%5E2%2Bu-1.02=0
Quadratic equation, you can solve by quadratic formula or completing the square.
u%5E2%2Bu%2B1%2F4-1.02=1%2F4
%28u%2B1%2F2%29%5E2=1%2F4%2B102%2F100
%28u%2B1%2F2%29%5E2=25%2F100%2B102%2F100
%28u%2B1%2F2%29%5E2=127%2F100
u%2B1%2F2=0+%2B-+sqrt%28127%29%2F10
u=-1%2F2+%2B-+sqrt%28127%29%2F10
So then,
sin%28theta%29=-1%2F2+%2B-+sqrt%28127%29%2F10
Since abs%28sin%28theta%29%29%3C1, then the value, -1%2F2-sqrt%28127%29%2F10 is not allowed since it is outside of this range.
.
.
.
sin%28theta%29=sqrt%28127%29%2F10-1%2F2
sin%28theta%29=0.6269
theta=38.8 and theta=141.2.
Both are in units of degrees.