Calculus Newton's Method

Question

I just need help on the second part where it asks to solve the equation exactly.

Approximate a solution of 6sin(𝑥)=7cos(2𝑥) in the interval [0,𝜋/2] to three decimal places, using Newton's method.

The solution is approximately .5522 .

 

Now solve the equation exactly.

The solution is ?

Answer 1

Remember that cos(2x) = 1 - 2sin²x

So your equation becomes:

6 sin(x) = 7[ 1 - 2sin²(x) ]

Distribute

6 sin(x) = 7 - 14 sin²(x)

Let u = sin(x)

6u = 7 - 14u²

Put everything on the left

14u² + 6u - 7 = 0

Using the quadratic formula find that

u = (-3/14) ± (√107 / 14)

So, replace each u with sin(x) [un-doing our substitution],

sin(x) = (-3/14) - (√107 / 14) or x = sin^-1 [ (-3/14) - (√107 /14) ] or

sin(x) = (-3/14) + (√107 /14) or x = sin^-1 [ (-3/14) + (√107 /14) ]

These each approximate to: -1.26347929 radians or -72.39203081 degrees

and 0.5522 radians or 31.63978316 degrees

But the inverse sine answers are the exact ones and are probably what your instructor wants.