Question

find the general solution of cos mx+sin nx=0​

Answer 1

Answer:

My attempt:

sin(mx)=−sin(nx)

=cos(π2−mx)=cos(π2+nx)

Using cosθ=cosα⇒θ=2nπ±α,

CASE 1:θ=2nπ+α

π2−mx=2pπ+(π2+nx)

⇒x=−2pπm+n

CASE 2:θ=2nπ−α

π2−mx=2qπ−(π2+nx)

⇒x=(2q−1)πn−m

⟹x=−2pπm+n or x=(2q−1)πn−m

Answer 2

cos mx +sin nx =0

cos mx+cos(90degree - nx)=0

=)cos(mx+90-nx)=0

=)mx+90+nx=0

=)mx+nx=-90

=)x(m+n)= -90

=)x= -90/m+n

thats solve