solve for
2cos2x=tan225
and,
solve using transformation method
cosx+ root2sinx= root3
THANKS!
tan 225 = tan 45 = 1; therefore
cos2x = 1/2
2x = 60 or 300 degreees
x = 30 or 150 degrees
I can't do the other one
for the first one, since we know tan 225 = 1
your equation is
2cos2x = 1
cos2x = 1/2
2x = 30 degrees or 2x = 150 degrees
x = 15 or x = 75 degrees
you state no domain, but since the period of cos2x is 180 degrees, more answers can be obtained by simply adding/subtracting multiple of 180 to the above answers,
e.g. 15 + 180 or 195 also works.
go with drwls solution, I was thinking sine
mine is wrong at the end
cosx+ sqrt2*sinx= sqrt3
Divide both sides by sqrt3
(1/sqrt3)*cosx + sqrt(2/3)*sinx = 1
Note that 1/sqrt3 = cos 54.74 degrees
and sqrt(2/3) = sin 54.74 degrees
Now use the "transformation" identity
cosa cosb + sina sinb = cos(a-b)
cos(a - 54.74deg)= 1
a -54.74 = 0 degrees
a = 54.74 deg
Check:
cosx = 0.5773
sqrt2*sinx = 1.1548
sum = 1.732 looks good
the second: ...
cosx + √2sinx = √3
√2sinx = √3 - cosx
square both sides
2sin^2 x = 3 - 2√3cosx + cos^2x
2(1 - cos^2x) = 3 - 2√3cosx + cos^2x
which simplifies to
3cos^2 x - 2√3cosx + 1 = 0
solving this quadratic for cosx yields
cos x = (2√3 ± √0)/6 = √3/3
x = 54.7356 degrees or 305.2644 degrees
since I squared, all answers have to be verified.
subbing back in, x = 54.7356 works, but x = 305.2644 does not work, so
x = 54.7356 degrees
I also did this using Calculus,
consider y = cosx + √2sinx
dy/dx = -sinx + √2cosx
= 0 for a max/min
then sinx = √2cosx
sinx/cosx = √2
tanx = √2
x = arctan √2 = 54.7356 degrees
subbing this back gave me 1.7321.. which is √3
What a way to spend Sunday morning, lol
... and on top of all I didn't even use the method of transformation that was asked for...
Oh well, some days .....
I'm glad we agreed.
I figured out the 'scale factor' needed change the coefficients on the left, to enable a "transformation", while scribbling on the back of a breakfast napkin.
I was thinking to use the fact that any linear combination such as
√2sinx + cosx can be written as
k(sin(x+T) , where T is the translation and k is the net amplitude.
k(sinxcosT + cosxsinT) = √2sinx + cosx
so
ksinxcosT = √2sinx and kcosxsinT= cosx
cosT = √2/k and sinT = 1/k
then tanT = 1/√2
T = 35.26 degrees, the angle in a 1: √2:3 triangle
then back in sinT = 1/k, k= √3
so k(sin(x+T) = √2sinx + cosx = √3
becomes
√3sin(x + 35.264) = √3
sin(x + 35.264) = 1
x + 35.264 = 90
x = 54.74
How is that for "overkill"
I liked the way you recognized the 1 : √2 : 3 ratio right away.
1024*(GT430)
To solve the equation 2cos2x = tan225, we need to find the value(s) of x that satisfy the equation. Here's how you can solve it:
Step 1: Simplify the equation.
Recall that tan(theta) can be expressed as sin(theta)/cos(theta). So, tan225 = sin225/cos225.
Step 2: Convert the angle from degrees to radians.
To work with trigonometric functions, we need to convert the angle from degrees to radians. Since 1 degree = π/180 radians, 225 degrees = 225π/180 radians.
Step 3: Substitute the simplified equation into the original equation.
Now, the equation becomes 2cos(2x) = sin(225π/180)/cos(225π/180).
Step 4: Simplify the right-hand side of the equation.
Using the values of sin(225π/180) and cos(225π/180) from the unit circle, the equation becomes 2cos(2x) = -1/√2.
Step 5: Solve for cos(2x).
To solve for cos(2x), divide both sides by 2. This gives us cos(2x) = -1/(2√2).
Step 6: Find the value(s) of x.
Using the identity cos(2x) = 1 - 2sin^2(x), we can rewrite cos(2x) = -1/(2√2) as 1 - 2sin^2(x) = -1/(2√2). Solve this equation for sin(x), and you will get two values for sin(x).
Step 7: Solve for x.
Once you have the values for sin(x), use the inverse sine function to find the values of x. Remember that the range of the inverse sine function is from -π/2 to π/2.
Now, let's move on to the second equation: cos(x) + √2sin(x) = √3.
This equation can be solved using the transformation method. Here's how:
Step 1: Square both sides of the equation.
(cos(x) + √2sin(x))^2 = (√3)^2.
Step 2: Expand and simplify.
cos^2(x) + 2√2sin(x)cos(x) + 2sin^2(x) = 3.
Step 3: Use the trigonometric identity sin^2(x) + cos^2(x) = 1 to simplify.
1 + 2√2sin(x)cos(x) + 2sin^2(x) = 3.
Step 4: Rearrange and divide by 2.
2sin^2(x) + 2√2sin(x)cos(x) - 2 = 0.
Step 5: Factorize.
sin^2(x) + √2sin(x)cos(x) - 1 = 0.
Now, you can solve this quadratic equation either by factoring, using the quadratic formula, or completing the square. Once you have the values for sin(x), you can use the inverse sine function to find the values of x.
Remember to always check if the values obtained satisfy the original equation.