Solving A Trigonometric Exponential Equation

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PLAYLISTS 🎵 :

[saidtakadda]

Hello,
i use the same method, i think the solutions are:

Pi/6 + k*Pi , -Pi/6 + k*Pi , Pi/3 + k*Pi , -Pi/3 + k*Pi where k belong to Z

[nizamkanchon]

Could u please kindly give me the pdf of aops introduction to algebra solutions manual?
Anyway, ur work has been a great help. Please keep it up

[quantumbuddha]

Fun problem! Thanks! I generalized the solution to (3k+/-1)pi/6, k€Z

[phill]

More steps is better, just watch at 2x

[pablocopello]

+-Pi/6 + k*Pi/2; k integer.
OR
+-(Pi/6 + n*Pi/2); n natural

[SidneiMV]

4^sin²x = u

u + 4/u = 3√2
u² - (3√2)u + 4 = 0

u = (3√2 ± √2)/2

u = 2√2 or u = √2

u = 2√2
4^sin²x = 2√2 = 2^(3/2)
2sin²x = 3/2 => sin²x = 3/4
sinx = ± (√3)/2

u = √2
4^sin²x = √2 = 2^(1/2)
2sin²x = 1/2 => sin²x = 1/4
sinx = ± 1/2

*x = kπ/6 (except for k = 3n)*