Question Number 10417 by okhema francis last updated on 08/Feb/17

hence or otherwise solve sin 6θ−sin 2θ=0 for 0≤θ≤(π/2).

Answered by nume1114 last updated on 08/Feb/17

    sin (α+β)−sin (α−β)  =(sin αcos β+cos αsin β)      −(sin αcos β−cos αsin β)  =2cos αsin β       α←4θ,β←2θ  ⇒sin 6θ−sin 2θ=2cos 4θsin 2θ         sin 6θ−sin 2θ=0  ⇔2cos 4θsin 2θ=0  ⇔cos 4θ=0 ∨ sin 2θ=0  cos 4θ=0⇔4θ=(π/2),((3π)/2)⇔θ=(π/8),((3π)/8)  sin 2θ=0⇔2θ=0,π⇔θ=0,(π/2)  ∴ θ=0,(π/8),((3π)/8),(π/2)

Commented byokhema francis last updated on 09/Feb/17

thank you and let the lord be with u

Answered by mrW1 last updated on 08/Feb/17

since sin 3x=3sin x−4sin^3  x  ⇒sin 6θ=3sin 2θ−4sin^3  2θ  3sin 2θ−4sin^3  2θ−sin 2θ=0  2sin 2θ(1−2sin^2  2θ)=0    ⇒sin 2θ=0  ⇒2θ=0 or π  ⇒θ=0 or (π/2)    ⇒1−2sin^2  2θ=0  ⇒sin 2θ=±(√(1/2))=±((√2)/2)  with 0≤θ≤(π/2), sin 2θ≥0  ⇒sin 2θ=((√2)/2)  ⇒2θ=(π/4) or ((3π)/4)  ⇒θ=(π/8) or ((3π)/8)    all solutions:  θ=(0,(π/8),((3π)/8),(π/2))

Commented byokhema francis last updated on 09/Feb/17

thank you guys for helping and let the lord go with you