Question

5.\(\left(\sin x+\frac{\sin3x+\cos3x}{1+2\sin2x}\right)=\cos2x+3\)

ai giải pt trên giúp tui với :(((

tui đang cần gấp

#mã mã#

Answer 1

ĐKXĐ: ...

\(5\left(\frac{sinx+2sin2x.sinx+sin3x+cos3x}{1+2sin2x}\right)=cos2x+3\)

\(\Leftrightarrow5\left(\frac{sinx+cosx-cos3x+sin3x+cos3x}{1+2sin2x}\right)=cos2x+3\)

\(\Leftrightarrow5\left(\frac{sinx+sin3x+cosx}{1+2sin2x}\right)=cos2x+3\)

\(\Leftrightarrow5\left(\frac{2sin2x.cosx+cosx}{1+2sin2x}\right)=cos2x+3\)

\(\Leftrightarrow5cosx=cos2x+3\)

\(\Leftrightarrow5cosx=2cos^2x+2\)

\(\Leftrightarrow2cos^2x-5cosx+2=0\Leftrightarrow cosx=\frac{1}{2}\)