Solving \sin^2 x+1=2x

Alvin Pugh

Alvin Pugh

Answered question

2022-01-29

Solving sin2x+1=2x

Answer & Explanation

nick1337

nick1337

Expert2022-01-31Added 777 answers

11+sin2(x)212x1 the function f:xsin2(x)+12x is continuous at [12,1] f(12)>0  f(1)<0 and f(x)=sin(2x)2<0By IVT, there is a unique solution α in ]12,1[α=limn+un withu0=1 andun+1=unf(un)f(un)This numerical method is known as Newton-Raphson.
karton

karton

Expert2022-01-31Added 613 answers

Once it is proved that f(x)=sin2(x)2x+1 has a unique real zero in the interval [0,π4], its numerical determination is simple since f(x) is a positive and convex function on (0,π4) (due to f′′(x)>0), hence by applying Newtons

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?