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Find all solutions to the equation
in the interval [0, 21). Enter the
solutions in increasing order.
sin 2x = 2 sin x
x = [?],[1]
TT
Remember: sin 20 = 2 sin 0 cos 0

Find All Solutions To The Equation In The Interval 0 21 Enter The Solutions In Increasing Order Sin 2x 2 Sin X X 1 TT Remember Sin 20 2 Sin 0 Cos 0 class=

Sagot :

Corsaquix

Answer:

[tex]x=0, \\x=\pi[/tex]

Step-by-step explanation:

Recall the trigonometric identity [tex]\sin 2x=2\sin x\cos x[/tex].

Therefore, given [tex]\sin 2x=2\sin x[/tex], rewrite the left side of the equation:

[tex]2\sin x\cos x=2\sin x[/tex]

Subtract [tex]\sin x[/tex] from both sides:

[tex]2\sin x\cos x-2\sin x=0[/tex]

Factor out [tex]2\sin x[/tex] from both terms on the left:

[tex]2\sin x(\cos x-1)=0[/tex]

We now have two cases:

[tex]\begin{cases}2\sin x=0, x=k\pi\text{ for }k\in \mathbb{Z}\\\cos x-1=0,x=k2\pi\text{ for }k\in \mathbb{Z}\end{cases}[/tex]

Since the problem stipulates that [tex]x[/tex] is in the interval [tex][0, 2\pi)[/tex], we have:

[tex]\text{For }x\in [0, 2\pi):\\\sin x=0\rightarrow \boxed{x=0, x=\pi}\\\cos x-1=0\rightarrow \boxed{x=0}[/tex]

Recall that square brackets mean inclusive and parentheses mean exclusive. Therefore, [tex]2\pi \notin [0, 2\pi)[/tex].

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