Given :
An equation, 2cos ß sin ß = cos ß .
To Find :
The value for above equation in (0, 2π ] .
Solution :
Now, 2cos ß sin ß = cos ß
2 sin ß = 1
sin ß = 1/2
We know, sin ß = sin (π/6) or sin ß = sin (5π/6) in ( 0, 2π ] .
Therefore,
Hence, this is the required solution.
Answer:
Distribute, then combine like terms, subtract 10 from both sides of the equation, simplify, divide both sides of the equation by the same term, and simplify to get your solution.
Step-by-step explanation:
1.) Distribute
6−2(4−5)=40
6−8+10=40
2.) Combine like terms
6a-8a+10=40
-2a+10=40
3.) Subtract 10 from both sides of the equation
-2a+10=40
-2a+10-10=40-10
4.) Simplify
subtract the numbers
-2a+10-10=40-10
-2a=40-10
subtract the numbers
-2a=40-10
-2a=30
-2a=30
5.) Divide both sides of the equation by the same term
-2a=30
6.) Simplify
cancel terms that are in both the numerator or denominator
a=
Divide the numbers
a=
a=-15
a=-15
7.) Solution
a=-15