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3 years ago

How do you solve that (2cosx+1)(2cosx-1)(2cos2x-1)=2cos4x+1Please help it's urgent!

Mathematics

2 answers:

Fofino [41]3 years ago

5 0

(2cos x + 1)(2cos x - 1)(2cos 2x - 1 ) = 2cos 4x + 1 has proven and solved.

<h3>Further explanation</h3>

We will solve and proving an equation related to trigonometric identity.

$\boxed{ \ (2cos \ x + 1)(2cos \ x - 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }$

Let (2cos x + 1) as Part-A, (2cos x - 1) as Part-B, and (2cos 2x - 1) as Part-C.

We multiply Part-A with Part-B.

Recall that $\boxed{ \ (a + b)(a - b) = a^2 - b^2 \ }$

Do it carefully.

$\boxed{ \ (4cos^2x - 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }$ ... Equation-1

We use one property of trigonometric identities, i.e.,

$\boxed{ \ cos \ 2x = 2cos^2x - 1 \ } \rightarrow multiply \ by \ 2 \ \rightarrow \boxed{ \ 2cos \ 2x = 4cos^2x - 2 \ }$

Prepare it like this.

$\boxed{ \ 2cos \ 2x = 4cos^2x - 1 - 1 \ } \rightarrow \boxed{ \ 4cos^2x - 1 = 2cos \ 2x + 1 \ }$ ... Equation-2

Equation-2 is substituted into Equation-1.

$\boxed{ \ (2cos \ 2x + 1)(2cos \ 2x - 1 ) = 2cos \ 4x + 1 \ }$

$\boxed{ \ 4cos{^2}2x - 1 = 2cos \ 4x + 1 \ }$

On the left side, use the same property earlier with a little processed into, $\boxed{ \ 4cos{^2}2x - 1 = 2cos \ 4x + 1 \ }$

$\boxed{ \ 2cos \ 4x + 1 = 2cos \ 4x + 1 \ }$

So it has proven and solved.

<h3>Learn more</h3>

A case about trigonometric identities brainly.com/question/1430645
How does the vertical acceleration at point a compare to the vertical acceleration at point c brainly.com/question/2746519
About vector components brainly.com/question/1600633

Keywords: how do you solve that, (2cosx + 1)(2cosx - 1)(2cos2x - 1) = 2cos4x + 1, trigonometric identity, property, proving

suter [353]3 years ago

3 0

On the left side, we can collapse some terms:

$(2\cos x+1)(2\cos x-1)=4\cos^2x-1$

Recall the double angle identity for cosine:

$\cos^2x=\dfrac{1+\cos2x}2$

So we have

$4\cos^2x-1=\dfrac{4(1+\cos2x)}2-1=2\cos2x+1$

With the same reasoning, we can collapse that side further:

$(2\cos2x+1)(2\cos2x-1)=4\cos^22x-1=\dfrac{4(1+\cos4x)}2-1=2\cos4x+1$

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Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$