sin2x = cos(3x -10), find tanx
sin= 90 - (3x -10)
2x = 90 -3x + 10
2x =100 -3x
2x + 3x = 100 - 3x + 3x
5x = 100
5x/5 = 100/5
x = 20
tanx sin/cos
tanx = sin40/cos50
tried using cos(3x) = 4cos^3 x - 3cosx and sin(3x) = 3sinx - 4sin^3x
but the cos10 and sin10 really messed things up.
So how about Newton's method?
let y = sin2x - cos(3x - 10)
dy/dx = 2cos2x + 3sin(3x-10)
looking at the graph:
https://www.wolframalpha.com/input/?i=solve++sin2x%3D++cos%283x+-10%29
let's start with x = 1
newx = x - (sin2x - cos(3x-10)) / (2cos2x + 3sin(3x -10)
and using a "good calculator" with memory storage ....
start with a guess of x = 1
x -- newx
1 1.055433864
1.055433864 __ 1.051890393
1.051890393 ___ 1.05749655
1.05749655 ___ 1.057522203
1.057522203 ___ 1.057522204 <---- correct to 8 decimal places
looks like another solution near x = 2
x -- newx
2 2.107109029
2.107109029 ___ 2.101660281
2.101660281 ___ 2.095289164
2.095289164 ___ 2.13728505
2.13728505 ___ 2.145633424
2.145633424 __ 2.146017541
2.146017541 ___ 2.146018366
one more step should do it correct to 8 decimals
get as many solutions as you want
sin2x = cos(3x-10degree), find tanx, significant figures
You didn't say you wanted the units were in degrees,
unless otherwise stated, at this level we work in radians.
My derivative is only valid in radians.
Wolfram's graph assumes the 10 is in radians
so you will have to change the 10° to radians, use (3x-π/18)
repeat the above iteration, change the answer back to degrees,
then take the tan of the x
good luck.
To find tanx, we need to use the given equation sin2x = cos(3x - 10).
First, let's simplify the right side of the equation. We know that cosθ = sin(90° - θ), so we can rewrite cos(3x - 10) as sin(90° - (3x - 10)).
Next, let's simplify sin(90° - (3x - 10)). We can expand the brackets to get sin(90° - 3x + 10).
Since sin(90°) = 1, we can substitute it in and get 1 - sin(3x - 80°).
Now, we have the equation sin2x = 1 - sin(3x - 80°).
To solve this equation, we'll use the identity sin2θ = 1 - cos2θ.
Replacing sin2x with 1 - cos2x, we get 1 - cos2x = 1 - sin(3x - 80°).
Next, let's apply the identity cos2θ = 1 - sin2θ.
Replacing cos2x with 1 - sin2x, we get 1 - sin2x = 1 - sin(3x - 80°).
Now, our equation becomes 1 - sin2x = 1 - sin(3x - 80°).
Since we have 1 on both sides, we can cancel them out, leaving us with -sin2x = -sin(3x - 80°).
Next, we'll divide both sides by -1 to get sin2x = sin(3x - 80°).
Using the double angle formula sin2θ = 2sinθcosθ, we can rewrite sin2x as 2sinxcosx.
Now, our equation becomes 2sinxcosx = sin(3x - 80°).
We know that sin(A - B) = sinAcosB - cosAsinB, so we can rewrite sin(3x - 80°) as sin3xcos80° - cos3xsin80°.
Our equation becomes 2sinxcosx = sin3xcos80° - cos3xsin80°.
Now, let's simplify sin3xcos80° - cos3xsin80°. The value of cos80° is approximately 0.1736, and sin80° is approximately 0.9848.
Replacing these values, we get sin3x * 0.1736 - cos3x * 0.9848.
Now, our equation becomes 2sinxcosx = 0.1736sin3x - 0.9848cos3x.
Since the left side of the equation contains sinx and cosx, and the right side contains sin3x and cos3x, we need to convert the right side to have sinx and cosx.
We'll use the double angle formulas sin2θ = 2sinθcosθ and cos2θ = cos²θ - sin²θ to do this.
Using sin2θ = 2sinθcosθ, we can rewrite sin3x as 2sinxcos²x.
Using cos2θ = cos²θ - sin²θ, we can rewrite cos3x as cos²x - sin²x.
Now, our equation becomes 2sinxcosx = 0.1736(2sinxcos²x) - 0.9848(cos²x - sin²x).
Simplifying further, we get 2sinxcosx = 0.3472sinxcos²x - 0.9848cos²x + 0.9848sin²x.
Moving all the terms to one side, we get 0 = 0.3472sinxcos²x - 0.9848cos²x + 0.9848sin²x - 2sinxcosx.
Factoring out a common term, we get 0 = sinx(cos²x(0.3472 - 2) + sin²x(0.9848 - 1)) - cosx(0.9848 - 0.3472cosx).
Simplifying further, we get 0 = sinx(-1.6528cos²x + 0.9848sin²x) - cosx(0.6376 - 0.3472cosx).
Now, we can see that this equation is quadratic in form, so we can solve it using quadratic techniques.
To find the values of x that satisfy this equation, we can apply the zero-product property, which states that if a * b = 0, then a = 0 or b = 0.
So we set each factor in the equation to 0 separately and solve for x.
First, sinx = 0:
To find the values of x where sinx = 0, we know that sinx = 0 when x is a multiple of π, so we can write x = nπ, where n is an integer.
Second, -1.6528cos²x + 0.9848sin²x = 0:
We can divide both sides by 0.9848 to simplify the equation: -1.6804cos²x + sin²x = 0.
Now, we can use the identity sin²x + cos²x = 1 to rewrite the equation:
1 - cos²x - 1.6804cos²x = 0.
Combining like terms and rearranging the equation, we get -2.6804cos²x = -1.
Dividing by -2.6804, we get cos²x = 1/2.6804.
Taking the square root of both sides, we get cosx = sqrt(1/2.6804).
Since cos(x) = sqrt(1 - sin²(x)), we can substitute sqrt(1 - sin²x) for cos(x):
sqrt(1 - sin²x) = sqrt(1/2.6804).
Squaring both sides to eliminate the square root, we get 1 - sin²x = 1/2.6804.
Simplifying the right side, we get 1 - sin²x = 0.3733.
Rearranging the equation, we get sin²x = 0.6267.
Taking the square root of both sides, we get sinx = sqrt(0.6267).
So we have two equations:
1. sinx = 0
2. sinx = sqrt(0.6267)
Solving equation 1, sinx = 0, we know that x = nπ, where n is an integer.
Solving equation 2, sinx = sqrt(0.6267), we can use the inverse sine function to find the values of x.
x = sin⁻¹(sqrt(0.6267)) + 2πn or x = π - sin⁻¹(sqrt(0.6267)) + 2πn, where n is an integer.
Now we have found the values of x that satisfy the equation sin2x = cos(3x - 10).
To find tanx, we can use the identity tanx = sinx/cosx.
So, the values of tanx are tan(nπ), sqrt(0.6267)/(sqrt(1/2.6804) + tan(sin⁻¹(sqrt(0.6267)) + 2πn), and sqrt(0.6267)/(sqrt(1/2.6804) - tan(sin⁻¹(sqrt(0.6267)) + 2πn)), where n is an integer.