Find all solutions to the equation sin2x+sinx-2cosx-1=0 in the interval [0, 2pi)
1 answer:
The solutions appear to be {π/2, 2π/3, 4π/3}.
_____
Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2
and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}
_____
Replacing sin(2x) with 2sin(x)cos(x), you have
2sin(x)cos(x) +sin(x) -2cos(x) -1 = 0
sin(x)(2cos(x) +1) -(2cos(x) +1) = 0 . . . . factor by grouping
(sin(x) -1)(2cos(x) +1) = 0
This has solutions
sin(x) = 1
x = π/2
and
2cos(x) = -1
cos(x) = -1/2
x = {2π/3, 4π/3}
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Answer:
<h2>Step-by-step explanation:
To find y when x=4, we must first find the relationship between them
The statement
y varies directly with x is written as
<h3>where k is the constant of proportionality
From the question
when y = 1/5
x = 3
Substitute the values into the above formula and solve
That's
<h3>So the formula for the variation is
<h3>From the question
when x = 4
<h3>Hope this helps you