how do you solve this problem using double angle formulas: sin2xsinx=cosx?

Answer 1

First, you must know this trig. identity:

sin[2x] = 2sin[x]cos[x]

Now, here's what you should do:

sin[2x]sin[x] = cos[x]

Substitute sin[2x] with 2sin[x]cos[x]:

2sin[x]cos[x]sin[x] = cos[x]

Multiply the two sin[x]:

2sin²[x]cos[x] = cos[x]

Divide both sides by cos[x]:

2sin²[x] = 1

Divide both sides by 2:

sin²[x] = ±√½

Take the square root of both sides:

sin[x] = ±(√2)/2

x = arcsin[±(√2)/2]

x = (π/4 ± 2πk), (3π/4 ± 2πk), (5π/4 ± 2πk) and (7π/4 ± 2πk)

... where 'k' is an natural number.

Answer 2

sin (2x) sin x = cos x

Use the double angle formula for sin:

2 sin x cos x sin x = cos x

Simplify:

2 sin² x cos x = cos x

We consider two cases. If cos x=0, then x=π/2+2πk or x=-π/2+2πk, which in this case may be combined into the single expression x=π/2+πk (where k is of course an arbitrary integer). Now if cos x≠0, we divide by it to obtain:

2 sin² x = 1
sin² x = 1/2
sin x = ±1/√2

So if sin x = 1/2, then x = π/4+2πk or x=3π/4+2πk. If sin x=-1/2, then x=-π/4+2πk or x=-3π/4+2πk. These expressions may be combined into the single expression x=π/4+πk/2, where k is an arbitrary integer. Therefore our entire solution set is {π/2+πk: k∈Z) ∪ {π/4+πk/2: k∈Z}.

Answer 3

it really is not a formula, it really is a Quadratic Equation, that you remedy through using the Quadratic formula. If the equation is ax^2 + bx + c = 0, then your 2 thoughts are x = (-b +/- sqrt(b^2 - 4ac))/2a. First rearrange your equation to get 2x^2 - (11/4)x - 35/4 = 0 - a minimum of, i assume it really is what it develop into meant to be. probable extra elementary now to multiply each little thing through 4 to get rid of the fractions, so 8x^2 - 11x - 35 = 0. So a = 8, b = - 11 and c = -35 so x = [11 +/- sqrt(121 + 4*8*35)]/16 = [11 +/- sqrt681]/16, so use your calculator. If really your first line develop into impressive themn we've 2x^2 - 1111/4 = 35/4; upload 111/4 to both side to get 2x^2 = 40 six/4, and dividing through 2 we get x^2 = 23/4 so take the suare root and get x = +/- (sqrt23)/2.

Answer 4

Use the identity sin(2θ) = 2sin(θ)cos(θ). So this becomes

(2sin(x)cos(x)) sin(x) = cos(x)

Right now it looks like we should divide both sides by cos(x), but this wouldn't work in the case of when cos(x) = 0. This is true when x = ±π/2, ±3π/2, ±5π/2 etc. So those are solutions. Now let's consider when cos(x) is NOT zero. Divide both sides by cos(x) to get
(2sin(x)) sin(x) = 1
2sin^2 (x) = 1
sin(x) = ±√(1/2) = ±√2/2
So x = ±π/4, ±3π/4, ±5π/4, etc. are solutions too.