1)Solve sin2x=3cosx for all values of x.
What do I need to do?
2)Solve e(sqrtx)=4
3)Approximate the greatest real root of f(x)=x^3-2x-5 to the nearest tenth.
I just need help not do for me. HELP. Please and Thank you. Explain please no links.
1: sin(2x) = 2sin(x)cos(x), so you know that 2sin(x)cos(x)=3cos(x). So either 2sin(x)=3, or cos(x)=0. You now need to find all the values of x such that one or other of these equations is true.
2: e^(sqrt(x))=4
so ln(e^(sqrt(x))=ln(4), but ln(e^y)) = y for any y,
so sqrt(x)=ln(4),
so x = (ln(4))^2. Try feeding this value into the original equation, and you should get 4.
3: Try to find any real root like this:
If x^3 - 2x - 5 = 0, then x^3 = 2x + 5, so x = (2x+5)^(1/3), i.e. the cube root of (2x+5).
Try x = 1: feed this into the equation and you'll get (2+5)^(1/3) = 1.913.
Now feed x=1.913 into the equation and you'll get 2.067.
Now feed x=2.067 into the equation and you'll get 2.090.
Now feed x=2.090 into the equation and you'll get 2.094.
Now feed x=2.094 into the equation and you'll still get 2.094 to 3 places of decimals.
The series is converging to 2.094, or 2.1 to one place of decimals, so this is one of the roots (though you don't know whether it's the highest one yet).
If you try it out, you should find that f(2.094)=0. (There are tests to enable you to find out whether the series will converge or not, but it's usually worth a try to see if you get lucky. If you don't get lucky, just try reformulating the equation another way and try that instead.) For x higher than this value, the gradient f'(x) = 3x² - 2 is always positive, so x=2.1 must be the greatest real root.
1) To solve sin(2x) = 3cos(x), we can use the trigonometric identity sin(2x) = 2sin(x)cos(x).
First, substitute sin(2x) with 2sin(x)cos(x):
2sin(x)cos(x) = 3cos(x).
Now, we can cancel out cos(x) from both sides:
2sin(x) = 3.
Next, divide both sides by 2:
sin(x) = 3/2.
Since the range of the sine function is -1 to 1, there are no values of x for which sin(x) equals 3/2. Therefore, the equation sin(2x) = 3cos(x) has no solutions.
2) To solve e(sqrt(x)) = 4, we need to isolate the square root term and then apply the natural logarithm (ln) to both sides.
First, raise both sides to the power of ln:
ln(e(sqrt(x))) = ln(4).
Using the property ln(e^a) = a, we can simplify the left side:
sqrt(x) = ln(4).
Now, square both sides to eliminate the square root:
(sqrt(x))^2 = (ln(4))^2.
Simplifying further, we have:
x = ln(4)^2.
Using a calculator, evaluate ln(4)^2 to find the exact value of x.
3) To approximate the greatest real root of f(x) = x^3 - 2x - 5 to the nearest tenth, we can use methods such as graphical analysis or numerical methods like the Newton-Raphson method.
For graphical analysis, you can plot the graph of f(x) and determine the x-coordinate of the highest point on the curve. This highest point corresponds to the greatest real root of the equation. Use a graphing tool or software to plot the graph and find the approximate x-coordinate.
For numerical methods like the Newton-Raphson method, you can iteratively find a solution using an initial guess. Starting with an initial guess, apply the Newton-Raphson iteration formula:
x1 = x0 - f(x0) / f'(x0),
where x0 is the initial guess, f(x0) is the function value at x0, and f'(x0) is the derivative of the function at x0.
Keep iterating this formula until you reach a value of x that satisfies the equation f(x) ≈ 0.
This method requires multiple iterations, and you can repeatedly apply the formula using a calculator or spreadsheet until you reach the desired degree of accuracy.