An Exponential Trigonometric Equation

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[@barberickarc3460]

Raise everything to the power of sin(x) and write the 16 as 4^2.
Bases are equal so exponents are equal, notice the left side is the same as sin(2x)
Everything simplifies to 2x = arcsin(1)
X = pi/4 + N*pi

[@honestadministrator]

4 ^ ( 2 cos ( x) - 1/ sin(x)) = 1
sin ( 2 x) = 1
2 x = ( 2 n + 1/2) π
x = ( n + 1/4) π

[@mikecaetano]

Apply the Pythagorean trigonometric substitution immediately to 1 and put all the terms on one side and you get (sinx-cosx)^2 = 0. [or (cosx-sinx)^2 it makes no difference]. This implies that sinx == cosx, which only occurs at pi/4 and 3 pi/4 with a period of pi.

[@scottleung9587]

I also got x=pi/4 + pi*n.

[@Sg190th]

2nd method is what I went for.

[@kassuskassus6263]

Instead of adding 1 to both sides, we can just put cos^2x+sin^2s-2cosxsinx=0, which is (cosx-sinx)^2=0, so cosx=sinx, then x=pi/4 +npi

[@michaeledwards2251]

Why not use , when 2 cos x sin x =1, the double angle formula 2 cos x sin x = cos 2x, so cos 2x = 1, Since cos (pi/2) = 1, x = pi/4 + n pi.

[@michaeledwards2251]

I should have said 2 cos x sin x = sin 2x, and sin 2x = 1, sin(pi/2) = 1

[@zygoloid]

4^(2cos x) = 4^csc x
2 cos x = 1/sin x
2 sin x cos x = 1 and x ≠ nπ
sin 2x = 1
2x = π/2 + 2nπ
x = π/4 + nπ

[@michaeledwards2251]

Why not use , when 2 cos x sin x =1, the double angle formula 2 cos x sin x = sin 2x, so sin 2x = 1, Since sin (pi/2) = 1, x = pi/4 + n pi.

[@We_Love_Basketball.]

x=π/4 + 2πn

[@giuseppemalaguti435]

2cosx=1/sinx...sin2x=1..x=45

[@vladimirkaplun5774]

sin 2x =1 is abusing distinguished professor?