I need help with calc..please?

Question

1. cos^2(x) - sin^2(x)=0 Domain for x is [0, 2pi]

2. x=Square root of x

3. 1/(2x-1) + 1/(5x+3)=0

4. xe^x - e^x=0 note that "e" is the base of natural logarithm, i.e 2.7182..

5. x^3 - 2x^2 - 5x + 6 = 0. Note that (x-1) is a factor of the equation.

6. sin^3 (x) - 2cos^3 (x)=0

Answer 1

1. You must write cos^2x in terms of sin^2x or vice versa:

This is easy because sin^2x + cos^2x = 1 ==> cos^2x = 1 - sin^2x

1-sin^2x - sin^2x = 0

1-2sin^2x = 0

Now factorize (dif betw squares): (1-√2sinx)(1+√2sinx) = 0

==> sinx = ±1/√2

==> x = π/4; 3π/4; 7π/4

2. x = √x ==> x² = (√x)² ==> x² = x ==> x² - x = 0

==> x(x-1) = 0 ==> x = 0 or x = 1

3. 1/(2x-1) + 1/(5x+3)

Use common denom: (5x+3 + 2x-1)/((2x-1)(5x+3)) = 0

==> 7x + 2 = 0....Only the numerator can be zero...

==> 7x = -2 ==> x = -2/7

4. Take the common factor e^x:

e^x(x-1) = 0 ==> x-1 = 0 ........... e^anything can never be zero...

Check this for yourself....remember e^0 = 1 and e^(-n) = 1/(e^n)

==> x = 1

5. (x-1) .... you are given this. Now think logically. You need to multiply

(x-1) by something that will give you x^3 - 2x^2 - 5x + 6.

So you can immediately write down (x-1)(x^2........-6)

The missing term is easy to find. What is more, you have two ways to do it. So you can choose one method to find it and the other to check.

You need something that when added to -6x will give you -5x. OR you need something that when added to -x^2 will give you -2x^2. Take a minute to see why this is so.

So, (x-1)(x^2 - x - 6) = 0

==> (x-1)(x+2)(x-3) = 0

==> x = 1, -2 or 3

6. sin^3x - 2cos^3x = 0

==> sin^3x = 2cos^3x

==> sinx = 2^(1/3)cosx....take cube root both sides...

==> sinx/cosx = 1/2^3

==> tanx = 1/8

==> x = 0.124354994 rad.