An Incredible Exponential Equation | Can You Find Real Solutions?

An Incredible Exponential Equation | Can You Find Real Solutions?
Welcome to infyGyan !! In this video, we'll be solving a nice exponential equation that's both challenging and fun. Perfect for algebra enthusiasts and students preparing for advanced math competitions, this tutorial will take you step-by-step through the process, helping you understand the concepts and techniques needed to tackle exponential equations with confidence.
Join us as we explore this intriguing problem and enhance your algebra skills. If you find the video helpful, don't forget to like, subscribe, and hit the bell icon for more exciting math challenges and tutorials.
Topics covered:
Exponential equation
Trigonometry
How to solve exponential equations?
Algebra
Properties of exponents
Cubic equation
Factorization
Algebraic identities
Trigonometric identities
Quadratic equation
Quadratic formula
Exponential Equation
Math Olympiad preparation
Math Olympiad training
Exponent laws
Real solutions
#exponentialequations #mathematics #math #matholympiad #problemsolving #algebra #trigonometry #educational
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Thanks for Watching !!

[@RashmiRay-c1y]

Let 2^(sin^x) = a. So, a is positive. Then, a + 4/a^2=3 > a^3-3a^2+4=0 > a= -1,2. But a cannot be -2. Thus, sin^x=1 > x = +/- pi/2 + 2 pi n, n being an integer.

[@dorkmania]

Evaluating in terms of power cos²(x) or ultimately converting the evaluated exponent in terms of cosine simplifies/reduces solution to cos(x) = 0 or x = π/2 + n•π

[@Shobhamaths]

Let 2^(cos^2(x)) =t,
2+t^3=3t;
t^3-3t+2=0;t=1, 1,-2
t=1; satisfies
cos^2(x) =0
x=π/2, 3π/2..odd multiples

[@michaeldoerr5810]

Umm...actually I think that the professor's answer is more precise because 3pi/2 is part of the second part of the full answer. And then again it is a trig function so it makes sense that the answers were represented in a certain way.

[@Shobhamaths]

@@michaeldoerr5810 yes 👍

[@RajeshKumar-wu7ox]

π÷2, 3π÷2

[@Fjfurufjdfjd]

Η σχεση γραφεται: y^3-3y^2+4=0 οπου y=2^(ημχ)^2=2^[(sin^2)χ] εχω(y+1)(y-2)^2=0 y=-1 απορριπτεται ή y=2 διπλη. (ημχ)^2=1 ημχ=1 ή ημχ=-1 ημχ=ημ(π/2) χ=2Κπ+(π/2) Κ€Z ;(σε rad) ή χ=360Κ +90 (σε μοιρες) ημχ=-1=ημ(-π/2) χ=2Κπ-π/2 ή χ=2ΚΠ+π-(-π/2)=2Κπ+3π/2(rad) ή χ=360Κ+270 (μοιρες) αρα τελικα χ=360Κ+90 ; χ=360Κ-90 ; χ=360Κ+270

[@Fjfurufjdfjd]

Αν και ξερω οτι δεν θα απαντησετε θελω να σας πω οτι το π=3.14159.... προερχεται απο το πρωτο γραμμα της ελληνικης λεξης περιφερεια που σημαινει κυκλος