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December 19, 2014

December 19, 2014

Posted by **bandit** on Wednesday, January 12, 2011 at 6:31pm.

solve the equation 1+25cos(x)=25sin^2(x)

- pre-calc -
**MathMate**, Wednesday, January 12, 2011 at 7:01pmSubstitute sin²(x)=1-cos²(x).

This results in a quadratic equation in cos(x). Solve for y=cos(x) and evaluate x=arccos(y) if -1≤y≤1.

- pre-calc -
**helper**, Wednesday, January 12, 2011 at 7:07pm1+25cos(x)=25sin^2(x)

-25 sin^2 x + 25 cos x + 1

-25 (1 - cos^2 x) + 25 cos x + 1

-25 + 25 cos^2 x + 25 cos x + 1

25 cos^2 x + 25 cos x - 24

substitute u = cos x

25u^2 + 25u - 24

complete the square

u^2 + u - 24/25 = 0

u^2 + u = 24/25

u^2 + u + 1/4 = 121/100

(u + 1/2)^2 = 121/100

take square root of both sides

| u + 1/2 | = 11/10

u + 1/2 = -11/10 and u + 1/2 = 11/10

u = -8/5 and u = 3/5

substitute back in u = cos x

cos x = -8/5 and cos x = 3/5

x = arccos (-8/5)

x = arccos (3/5)

- pre-calc -
**helper**, Wednesday, January 12, 2011 at 7:15pmforgot to put

I am not a tutor

I think this is right

- pre-calc -
**MathMate**, Wednesday, January 12, 2011 at 8:22pmIt's almost perfect.

In the last step,

"x = arccos (-8/5)

x = arccos (3/5) "

we have to reject the first value of arccos(-8/5) because arccos(-8/5) does not exist.

For a solution between 0 and 2π, there are two values of x that satisfy arccos(3/5), i.e.

approx. 53 degrees and 360-53 degrees.

For the general solution, add 2kπ to the solutions, where k is an integer.

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