We know that sin2x=2sinxcosx
(search the net for proof if you wish)
So the original equation becomes
2sinxcosx-sinx=0
The two terms both have sinx that can be taken out to get:
sinx(2cosx-1)=0
This is true if sinx=0 or 2cosx-1=0 , rewritten: cosx=1/2
sinx=0 than x=2kπ
cosx=1/2 than x=π/3+2kπ
where k is an integer
Lets get these into improper fractions.
1*8 + 5 = 8+5 = 13
13/8
1*3 + 2 = 3+2 = 5
5/3
Now divide
13/8 = 1.625
5/3 = 1.6666
1 2/3 is larger
The answer should be AC+BC
We can answer this by applying the rule called "Greater angle, greater side". This is where in any triangle, the opposite side of the larger angle is also longer.
So in this question, first we have to find the largest 2 angles, which is ∠ABC and ∠BAC with measures of 120° and 35° respectively. Therefore, from the diagram, the opposite sides of the angles are AC and BC.
Therefore, the answer should be AC+BC.
Answer: C) x = 2
2^{2x + 2} = 2^{3x}
Since both terms (above) have the same base, set the exponents to be equal:
2x + 2 = 3x (Rearrange to solve for x)
x = 2
∴ x = 2
