Answer:
sin
(
x/
2
) = -
√
3
/2
Take the inverse sine of both sides of the equation to extract x
from inside the sine.
x/
2
=
arcsin
(
−
√
3/
2
)
The exact value of arcsin
(
−
√
3
/2
) is −
π
/3
.
/x
2
=
−
π
/3
Multiply both sides of the equation by 2
.
2
⋅
x
/2
=
2
⋅
(
−
π
/3
)
Simplify both sides of the equation.
x
=
−
2
π
/3
The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from 2
π
, to find a reference angle. Next, add this reference angle to π to find the solution in the third quadrant.
x
/2
=
2
π
+
π/
3
+
π
Simplify the expression to find the second solution.
x
=
2
π
/3
4
π
Add 4
π to every negative angle to get positive angles.
x
=
10
π
/3
The period of the sin
(
x
/2
) function is 4
π so values will repeat every 4
π radians in both directions.
x
=2
π
/3
+
4
π
n
,
10
π/
3
+
4
π
n
, for any integer n
Exclude the solutions that do not make sin
(
x
/2
)
=
−
√
3/
2 true.
x
=
10
π
/3
+
4
π
n
, for any integer n
Answer:
The slope of problem 11 is
The slope of problem 10 is
3
The 3 is the gradient in front of the x
Mark brainliest please
Angle C equals 63
Add angle 17 plus 100 and the subtract the answer from 180. Because a triangles are always 180 degrees
