Find all solutions to the equation sin(3x)cosx-sinx cos(3x)=0 on the interval [0,2pi]

a- x=0,pi/4,pi/2,3pi/4,pi,3pi/2,2pi
b- x=0,pi/2,pi,3pi/2,2pi
c- x=0,pi,2pi
d- x=0,pi/2,3pi/2

Cos(x)Cos(3x) + sin(x)sin(3x)=0 [0,2pi]

x= pi/4, 3pi/4, 5pi/4, 7pi/4

explanation:For those who have a different answer choice along with the question.

it can be to see completed examples before writing, in order to examine what makes for an effective piece. see how these intros give some context before posing the topic and lead naturally into the body paragraphs, where the evidence will be listed. you can imagine how the author has organized the rest of the piece.

"we buried my cousin last summer. he was 32 when he hanged himself from a closet coat rack in the throes of alcoholism, the fourth of my blood relatives to die prematurely from this deadly disease. if america issued drinking licenses, those four men—including my father, who died at 54 of liver failure—might be alive today. that is a good example piece   some steps that are .

. writing something that happened to you

.read a book you like to get ideas from there

. lastly you can make a venn diagram

hope this plz make brainliest

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.