Find all solutions in the interval [0, 2π).

2 sin2x = sin x

Answers:
x = π/3, 2π/3
x = π/2, 3π/2, π/3, 2π/3
x = 0, π, π/6, 5π/6
x= π/6, 5π/6

2sin^2(x) = sinx
we switch sinx from left to right and take sinx in front of brackets.

sinx*(2sinx - 1) = 0

now we need to solve 2 cases.
first:

sinx = 0
that is for x = 0 and x = pi

second:
2sinx - 1 = 0
2sinx = 1
sinx = 1/2
x = pi/6 or 5pi/6

the case we had here is something like a*b=0 so either a=0 or b=0 thats why we solved problem this way.

At the end comparing our results with options we can see that answer is third option.

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fqc5i7tsrt fqc5i7tsrt · Answer 2 · 2020-12-17T18:28:42+01:00

fqc5i7tsrt fqc5i7tsrt

17-12-2020

Answer:

It's x = 0, π, π/6, 5π/6

Step-by-step explanation:

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Find all solutions in the interval [0, 2π). 2 sin2x = sin x Answers: x = π/3, 2π/3 x = π/2, 3π/2, π/3, 2π/3 x = 0, π, π/6, 5π/6 x= π/6, 5π/6

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Find all solutions in the interval [0, 2π).

2 sin2x = sin x

Answers:
x = π/3, 2π/3
x = π/2, 3π/2, π/3, 2π/3
x = 0, π, π/6, 5π/6
x= π/6, 5π/6