Solve:
sin3x+cos2x=0
Given : sin3x+cos2x=0
Let us simplify,
⇒cos2x=-sin3x
⇒cos2x=–cos(π2–3x)[Since,sinA=cos(π2–A)]
⇒cos2x=cos(π–(π2–3x))[since,-cosA=cos(π–A)]
⇒cos2x=cos(π2+3x)
⇒ 2x=2nπ±(π2+3x)
⇒ 2x=2nπ+(π2+3x)or2x=2nπ–(π2+3x)
⇒3x-2x=-π2–2nπor2x+3x=2nπ–π2
⇒ x=-π2–2nπor5x=2nπ–π2
⇒ x=-π21+4nor5x=π24n–1
⇒ x=-π24n+1orx=π104n–1
Hence, The solution of the equation is x=-π24n+1orx=π104n–1 where n∈Z