Question

i need step by step instrustions on what to do when asked to find the exact solutions of the equation in the interval 0,2pi when the equation is 4sin x cos x

Answer 1

4sinx cosx =1

Answer 2

It's a tougher problem...but here's what you do:

Answer 3

i need to figure out how to do these for my test tomorrow

Answer 4

sin (2x) = 2 sin x cos x (that's a double-angle formula)

So, 4 sin x cos x is just 2 sin(2x)

because 2 sin(2x) = 2(2 sin x cos x) = 4 sin x cos x

So rewrite the equation as 2 sin(2x) = 1
Now, divide both sides by 2, you get

sin(2x) = 1/2

The question is..the sine of what angle is 1/2...

Answer many angles, like pi/6, 5pi/6, etc..

So take the angle 2x abd set is to pi/6

Answer 5

2x = pi/6

x = pi/12...so pi/12 is one solution.

Now, set 2x = 5pi/6

so solving for x, you get x = 5pi/12

so 5pi/12 is another solution to the problem.

Answer 6

continue this process till you find all solutions in (0, 2pi).

Answer 7

ahh ok. i see

Answer 8

that wopuld be all solutions if i believ so right?

Answer 9

You had to realize that 4 sin x cos x = 2 sin(2x)

Answer 10

No, there are more solutions than pi/12 and 5pi/12. You will find the remaining solutions...

Answer 11

i got to 2(2 sinx cosx) but didnt know what to do after that

Answer 12

okay

Answer 13

7pi over 12 and 11pi/12 is the last two right?

Answer 14

sekar is this right?