Respuesta :
[tex]\sin3\theta\cos\theta-\cos3\theta\sin\theta=\sin(3\theta-\theta)=\sin2\theta=\dfrac{\sqrt2}2[/tex]
[tex]\sin2\theta=\dfrac1{\sqrt2}[/tex]
[tex]\implies2\theta=\dfrac\pi4+2n\pi,\,2\theta=\dfrac{3\pi}4+2n\pi[/tex]
[tex]\implies\theta=\dfrac\pi8+n\pi,\,\theta=\dfrac{3\pi}8+n\pi[/tex]
where [tex]n[/tex] is any integer. To take only the solutions within the interval [tex]0\le\theta<2\pi[/tex], we solve
[tex]0\le\dfrac\pi8+n\pi<2\pi\implies\dfrac18+n<2\implies n<\dfrac{15}8\implies n=0,\,n=1[/tex]
[tex]\implies\theta=\dfrac\pi8,\,\theta=\dfrac\pi8+\pi=\dfrac{9\pi}8[/tex]
[tex]0\le\dfrac{3\pi}8+n\pi<2\pi\implies \dfrac38+n<2\implies n<\dfrac{13}8\implies n=0,\,n=1[/tex]
[tex]\implies\theta=\dfrac{3\pi}8,\,\theta=\dfrac{11\pi}8[/tex]
[tex]\sin2\theta=\dfrac1{\sqrt2}[/tex]
[tex]\implies2\theta=\dfrac\pi4+2n\pi,\,2\theta=\dfrac{3\pi}4+2n\pi[/tex]
[tex]\implies\theta=\dfrac\pi8+n\pi,\,\theta=\dfrac{3\pi}8+n\pi[/tex]
where [tex]n[/tex] is any integer. To take only the solutions within the interval [tex]0\le\theta<2\pi[/tex], we solve
[tex]0\le\dfrac\pi8+n\pi<2\pi\implies\dfrac18+n<2\implies n<\dfrac{15}8\implies n=0,\,n=1[/tex]
[tex]\implies\theta=\dfrac\pi8,\,\theta=\dfrac\pi8+\pi=\dfrac{9\pi}8[/tex]
[tex]0\le\dfrac{3\pi}8+n\pi<2\pi\implies \dfrac38+n<2\implies n<\dfrac{13}8\implies n=0,\,n=1[/tex]
[tex]\implies\theta=\dfrac{3\pi}8,\,\theta=\dfrac{11\pi}8[/tex]
Answer: For 0 ≤Ф≥ 2π (where π= 180°)
∴ Ф = 22.5°, 67.5°, 112.5°, 157.5°, 202.5°, 247.5°, 292.5°, 337.5°
Step-by-step explanation:
sin(3Ф)cos(Ф) - cos(3Ф)sin(Ф) = √2/2
sin(3Ф - Ф) =√2/2
3Ф -Ф = sin∧-1{√2/2}
2Ф = 45°
∴ Ф = 22.5°