A Trigonometric Exponential Equation

[@MrLidless]

Expand the sin2x, then take the cos root of both sides to 2^(2sinx) = 7, but the max the LHS can be is 4, so no real solutions when cosx0

[@RexxSchneider]

For the solutions where cosx is not zero:
sinx = log7/log4 (for any reasonable base of logarithms)
We can set sinx = (e^(ix) - e^(-ix)) / 2i.
Let y = e^(ix), so x = -i.ln(y)
Now sinx = (y - 1/y) / 2i
So (y - 1/y) / 2i = log7 / 2log2
y - 1/y = i.log7/log2
Let c = log7/2log2 = log7/log4
So y - 1/y = 2ic
y^2 - 2icy - 1 = 0
y = (2ic ± √(-4c^2 + 4))/2
y = ic ± √(-c^2 + 1) = ic ± i√(c^2 - 1) = i(c ± √(c^2 - 1)
x = -i.ln( i.(c ± √(c^2 - 1)) ) where c = log7/log4
You were probably right to use Wolfram Alpha.

[@scottleung9587]

Whoopsies, I cancelled out the cosines!

[@antenym8947]

I did that and it didnt work bruuuuh.

[@goldfing5898]

Addition theorem: sin(2x) = 2 * sin(x) * cos(x), so we can rewrite
2^(2*sin(x)*cos(x)) = 7^cos(x)
(((2^2)^sin(x))^cos(x) = 7^cos(x)
(4^sin(x))^cos(x) = 7^cos(x)
Exponent comparison:
4^sin(x) = 7
But is this unique?
sin(x) = log_4_(7)
To be continued...

[@georgesbv1]

Or cos(x) =0

[@goldfing5898]

@@georgesbv1 Yes, indeed.

[@rakenzarnsworld2]

@@goldfing5898 I literally done this for nothing other than finding cos(x) = 0 lol
log 2 = 0.301
log 7 = 0.8451
sin(2x) = 0.8451
cos(x) = 0.301
Let me do some approximations:
sin(1/36)pi = 0.0872
sin(2/36)pi = 0.1736
sin(3/36)pi = 0.2588
sin(4/36)pi = 0.342
cos(x) is larger than sin(3/36)pi but smaller than sin(4/36)pi
sin(5/36)pi = 0.4226
sin(6/36)pi = 0.5
sin(7/36)pi = 0.5736
sin(8/36)pi = 0.6428
sin(9/36)pi = 0.7071
sin(10/36)pi = 0.766
sin(11/36)pi = 0.8192
sin(12/36)pi = 0.866
sin(2x) is larger than sin(11/36)pi but smaller than sin(12/36)pi
sin(12/36)pi = sin(24/36)pi, sin(11/36)pi = sin(25/36)pi
cos(n/36)pi = sin{(n+18)/36}pi
sin{x+(18/36)pi} = 0.301
cos(12/36)pi = 0.5
sin(37/36)pi = -0.0872
sin(38/36)pi = -0.1763
...
sin(50/36)pi = -0.9397
sin(51/36)pi = -0.9659
sin(52/36)pi = -0.9848
sin(53/36)pi = -0.9962
sin(54/36)pi = -1
cos(21/36)pi = -0.2588
cos(22/36)pi = -0.342
...
cos(50/36)pi = -0.342
cos(51/36)pi = -0.2588
cos(50/36)pi = cos(122/36)pi = cos(94/36)pi, cos(51/36)pi = cos(123/36)pi = cos(93/36)pi
sin(48/36)pi = -sin(12/36)pi, sin(47/36)pi = -sin(11/36)pi
sin^2(x)+cos^2(x)=1
sin^2(2x)+cos^2(2x)=1
0.8451^2 = 0.71419401
0.301^2 = 0.090601
log2(7) = 2.8074
sin(2x) = 2.8074sin{x+(18/36)pi}
2cos(x)sin(x) = 2.8074cos(x)
cos(x)sin(x) = 1.4037cos(x)
sin(x) = 1.4037
However, sin(x) cannot be larger than 1
sin(2x) = cos(x) = 0
sin(x) = 0
x = 1.5708 + (n)pi
n is integer

[@gergelymadar4647]

sin(2x)=2cos(x)sin(x).
From here, we have 4^sin(x)cos(x)=7^cos(x).
(4^sin(x))^cos(x)=7^cos(x).
If cos(x)=0, x=1/2π+kπ, which is a solution, so we have one solution.
4^sin(x)=7
Log⁴(7)>|1|, so we have no more solutions. Hence, x=1/2π+kπ

[@carly09et]

Using log base 4 is neat.

[@juandavidjurado3738]

Thanks 👍🏻

[@seanfraser3125]

Using sin(2x) = 2sinxcosx, we have
4^(sinx*cosx) = 7^cosx
Taking the log of both sides we get
sinx*cosx*log4 - cosx*log7 = 0
Or,
cosx(log4sinx - log7) = 0
One family of solutions is when cosx = 0, or when x = pi/2 + pi*n for any integer n.
The other factor gives sinx = log7/log4 = log_4(7) > 1. But sinx can never be larger than 1, so this gives no (real) solutions.
So the only real solutions are the zeroes of cosx, or the odd integer multiples of pi/2
EDIT: Side note, I didn’t specify which base of log I used. Notice that it doesn’t matter as long as you choose one which is well-defined, since we eventually take a ratio anyway.

[@rakenzarnsworld2]

log 2 = 0.301
log 7 = 0.8451
sin(2x) = 0.8451
cos(x) = 0.301
Let me do some approximations:
sin(1/36)pi = 0.0872
sin(2/36)pi = 0.1736
sin(3/36)pi = 0.2588
sin(4/36)pi = 0.342
cos(x) is larger than sin(3/36)pi but smaller than sin(4/36)pi
sin(5/36)pi = 0.4226
sin(6/36)pi = 0.5
sin(7/36)pi = 0.5736
sin(8/36)pi = 0.6428
sin(9/36)pi = 0.7071
sin(10/36)pi = 0.766
sin(11/36)pi = 0.8192
sin(12/36)pi = 0.866
sin(2x) is larger than sin(11/36)pi but smaller than sin(12/36)pi
sin(12/36)pi = sin(24/36)pi, sin(11/36)pi = sin(25/36)pi
cos(n/36)pi = sin{(n+18)/36}pi
sin{x+(18/36)pi} = 0.301
cos(12/36)pi = 0.5
sin(37/36)pi = -0.0872
sin(38/36)pi = -0.1763
...
sin(50/36)pi = -0.9397
sin(51/36)pi = -0.9659
sin(52/36)pi = -0.9848
sin(53/36)pi = -0.9962
sin(54/36)pi = -1
cos(21/36)pi = -0.2588
cos(22/36)pi = -0.342
...
cos(50/36)pi = -0.342
cos(51/36)pi = -0.2588
cos(50/36)pi = cos(122/36)pi = cos(94/36)pi, cos(51/36)pi = cos(123/36)pi = cos(93/36)pi
sin(48/36)pi = -sin(12/36)pi, sin(47/36)pi = -sin(11/36)pi
sin^2(x)+cos^2(x)=1
sin^2(2x)+cos^2(2x)=1
0.8451^2 = 0.71419401
0.301^2 = 0.090601
log2(7) = 2.8074
sin(2x) = 2.8074sin{x+(18/36)pi}
2cos(x)sin(x) = 2.8074cos(x)
cos(x)sin(x) = 1.4037cos(x)
sin(x) = 1.4037
However, sin(x) cannot be larger than 1
sin(2x) = cos(x) = 0
sin(x) = 0
x = 1.5708 + (n)pi
n is integer

[@VVv-ix2gx]

1.5708 suspiciously looks like Pi/2

[@samvelhovhannisyan795]

x=pi/2 + pi*n

[@wearehumans3734]

Take, sin2x = cosx,
2sinxcosx=cosx
2sinx=1
Sinx=½
X=30°

[@neuralwarp]

You mis-spelt 7

[@broytingaravsol]

x=(4k±1)π/2

[@georgesbv1]

that logarithm is larger than 1

[@broytingaravsol]

@@georgesbv1 i've amended

[@giuseppemalaguti435]

Ho suluzioni per cosx=0 e per sinx=log(4)7 no real