Question

cos2x+sin^2x=0 на промежутке (0;3п) помогите ((((

Answer 1

cos2x + sin²x = 0    на промежутке (0; 3π)

$cos~2x+sin^2x=0\\ cos^2x-sin^2x+sin^2x=0\ \ cos^2x=0~~~Leftrightarrow~~~ cos~x=0\ \ boldsymbol{x=dfrac{pi }{2}+pi n,~~nin Z}$

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$n=-1;~~~x_0=dfrac{pi }{2}-pi=-0,5pi;~~~x_0notin (0;3pi )\ \ n=0;~~~x_1=dfrac{pi }{2}=0,5pi;~~~x_1in (0;3pi )\ \ n=1;~~~x_2=dfrac{pi }{2}+pi=1,5pi;~~~x_2in (0;3pi )\ \ n=2;~~~x_3=dfrac{pi }{2}+2pi=2,5pi;~~~x_2in (0;3pi )\ \n=3;~~~x_4=dfrac{pi }{2}+3pi=3,5pi;~~~x_4notin (0;3pi )\ \ \ boxed{boldsymbol{~~x_1=0,5pi;~~~x_2=1,5pi;~~~x_3=2,5pi ~~}}$