What is the general solution for 4e^{cos(x)}=3 ?

The general solution for 4e^{cos(x)}=3 is x=1.86260…+2pin,x=-1.86260…+2pin

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4e^{cos(x)}=3

4 e^{c o s (x)} = 3

x = 1.86260 \dots + 2 πn, x = - 1.86260 \dots + 2 πn

Degrees

x = 106.71923 \dots^{\circ} + 36 0^{\circ} n, x = - 106.71923 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 e^{c o s (x)} = 3

Divide both sides by

4

4 e^{c o s (x)} = 3

Divide both sides by

4

\frac{4 e ^{c o s (x)}}{4} = \frac{3}{4}

Simplify

e^{c o s (x)} = \frac{3}{4}

e^{c o s (x)} = \frac{3}{4}

Apply exponent rules

e^{c o s (x)} = \frac{3}{4}

f (x) = g (x)

, then

ln (f (x)) = ln (g (x))

ln (e^{c o s (x)}) = ln (\frac{3}{4})

Apply log rule:

ln (e^{a}) = a

ln (e^{c o s (x)}) = cos (x)

cos (x) = ln (\frac{3}{4})

cos (x) = ln (\frac{3}{4})

Apply trig inverse properties

cos (x) = ln (\frac{3}{4})

General solutions for

cos (x) = ln (\frac{3}{4})

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

x = arccos (ln (\frac{3}{4})) + 2 πn, x = - arccos (ln (\frac{3}{4})) + 2 πn

x = arccos (ln (\frac{3}{4})) + 2 πn, x = - arccos (ln (\frac{3}{4})) + 2 πn

Show solutions in decimal form

x = 1.86260 \dots + 2 πn, x = - 1.86260 \dots + 2 πn

- 2 sin (B) - 6 = sin (B) - 3

sin (x) (2 sin (x) + 1) = 2

4 sin (θ) + 1 = 0

sin^{2} (θ) (1 + csc^{2} (θ)) = k

- 3 cos (2 θ) + 9 cos (θ) - 7 = 4 cos (θ) - 3

What is the general solution for 4e^{cos(x)}=3 ?
The general solution for 4e^{cos(x)}=3 is x=1.86260…+2pin,x=-1.86260…+2pin