The good lady did not get all the solutions
between 0 and . There are 6 solutions for
the first one and 4 solutions for the second one.
How can i solve these trigonometric functions?
Solve for x∈[0,2π]:
The sine is negative in QIII and QIV
Rationalize the denominator of
and get and from your knowledge
of special angles or from the unit circle,
the solutions for 3x are 225° and 315° or
since you are using radians, and .
However, the left side is sin(3x), not sin(x).
That means we must get 3 times as many answers for x
as we would get for 3x.
To do that we add 3-1 = 2 times to each of those
answers for 3x in order to get all non-negative answers for
x less than :
So
1. we have to add and to the first answer:
which we get LCDs and simplify as
and simplify further as
Then divide through by 3 (multiply through by and get:
or
Also,
2. we have to add and to the second answer:
We get the LCD:
and simplify further as
Then divide through by 3 (multiply through by ) and get:
or
All the 6 solutions in order of magnitude are
------------------------
Solve for x∈[0,2π]:
As before, the sine is negative in QIII and QIV.
From your knowledge of special angles or from the unit
circle, the solutions for sin(2x) are 210° and 330° or
since you are using radians, and .
However, the left side is sin(2x), not sin(x).
That means we must get 2 times as many answers for x
as we would get for 2x.
To do that we add 2-1 = 1 time to each of those
answers for 2x in order to get all non-negative answers for
x less than :
So
1. we have to add to the first answer:
We get the LCD:
and simplify further as
Then divide through by 2 (multiply through by ) and get:
or
Also,
2. we have to add to the second answer :
which we get LCDs and simplify as
and simplify further as
Then divide through by 2 (multiply through by and get:
or
All the 4 solutions in order of magnitude are
Edwin