Answer:
[tex]x=n\pi[/tex] or [tex]x= \frac{(2n\pm1)\pi}{2}[/tex], where [tex]n\ge0[/tex]
Step-by-step explanation:
The given trigonometric equation is:
[tex](\sin x)(\cos x)=0[/tex]
By the zero product principle;
Either [tex]\sin x=0[/tex] or [tex]\cos x=0[/tex]
When [tex]\sin x=0[/tex], then [tex]x=n\pi[/tex]
When [tex]\cos x=0[/tex], then [tex]x=2n\pi \pm \cos^{-1}(0)[/tex]
This implies that: [tex]x=2n\pi \pm \frac{\pi}{2}[/tex]
Therefore the general solution is [tex]x=n\pi[/tex] or [tex]x= \frac{(2n\pm1)\pi}{2}[/tex], where [tex]n\ge0[/tex]
Answer:
B. pi divided by two plus n pi comma n pi such that n equals zero, plus or minus one, plus or minus two to infinity
Step-by-step explanation:
Given equation,
[tex](sin x)(cosx)=0[/tex]
[tex]\implies sinx=0\text{ or }cosx=0[/tex]
[tex]x=sin^{-1}(0)\text{ or }x=cos^{-1}(0)][/tex]
[tex]x=\pi, 2\pi, 3\pi,......\text{ or }x=\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}....[/tex]
[tex]\implies x = n\pi\text{ or }x=\frac{\pi+2n\pi}{2}[/tex]
Or
[tex]x=n\pi\text{ or }x= \frac{\pi}{2}+n\pi[/tex]
Where, n = 0, 1, -1, -2, ........∞
Hence, option 'B' is correct.
