What is the general solution for 2^{arccos(x)}+4^{arccos(x)+1}-2=3 ?

The general solution for 2^{arccos(x)}+4^{arccos(x)+1}-2=3 is x=1

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Popular Trigonometry >

2^{arccos(x)}+4^{arccos(x)+1}-2=3

Solution

2^{a r c c o s (x)} + 4^{a r c c o s (x) + 1} - 2 = 3

Solution

x = 1

Solution steps

2^{a r c c o s (x)} + 4^{a r c c o s (x) + 1} - 2 = 3

Solve by substitution

2^{a r c c o s (x)} + 4^{a r c c o s (x) + 1} - 2 = 3

Let:

arccos (x) = u

2^{u} + 4^{u + 1} - 2 = 3

2^{u} + 4^{u + 1} - 2 = 3 : u = 0

2^{u} + 4^{u + 1} - 2 = 3

Apply exponent rules

2^{u} + 4^{u + 1} - 2 = 3

Convert to base

2 : 2^{u} + 2^{2 (u + 1)} - 2 = 3

Convert

4

to base

2

4 = 2^{2}

2^{u} + (2^{2})^{u + 1} - 2 = 3

Apply exponent rule:

(a^{b})^{c} = a^{b c}

(2^{2})^{u + 1} = 2^{2 (u + 1)}

2^{u} + 2^{2 (u + 1)} - 2 = 3

2^{u} + 2^{2 (u + 1)} - 2 = 3

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

2^{2 (u + 1)} = 2^{2 u} \cdot 2^{2}

2^{u} + 2^{2 u} \cdot 2^{2} - 2 = 3

Apply exponent rule:

a^{b c} = (a^{b})^{c}

2^{2 u} = (2^{u})^{2}

2^{u} + (2^{u})^{2} \cdot 2^{2} - 2 = 3

2^{u} + (2^{u})^{2} \cdot 2^{2} - 2 = 3

Rewrite the equation with

2^{u} = v

v + (v)^{2} \cdot 2^{2} - 2 = 3

Solve

v + v^{2} \cdot 2^{2} - 2 = 3 : v = 1, v = - \frac{5}{4}

v + v^{2} \cdot 2^{2} - 2 = 3

Expand

v + v^{2} \cdot 2^{2} - 2 : v + 4 v^{2} - 2

v + v^{2} \cdot 2^{2} - 2

2^{2} = 4

= v + 4 v^{2} - 2

v + 4 v^{2} - 2 = 3

Move

3

to the left side

v + 4 v^{2} - 2 = 3

Subtract

3

from both sides

v + 4 v^{2} - 2 - 3 = 3 - 3

Simplify

4 v^{2} + v - 5 = 0

4 v^{2} + v - 5 = 0

Solve with the quadratic formula

4 v^{2} + v - 5 = 0

Quadratic Equation Formula:

For

a = 4, b = 1, c = - 5

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 4 ( - 5 )}{2 \cdot 4}

1^{2} - 4 \cdot 4 (- 5) = 9

1^{2} - 4 \cdot 4 (- 5)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 4 (- 5)

Apply rule

- (- a) = a

= 1 + 4 \cdot 4 \cdot 5

Multiply the numbers:

4 \cdot 4 \cdot 5 = 80

= 1 + 80

Add the numbers:

1 + 80 = 81

= 81

Factor the number:

81 = 9^{2}

= 9^{2}

Apply radical rule:

9^{2} = 9

= 9

v_{1, 2} = \frac{- 1 \pm 9}{2 \cdot 4}

Separate the solutions

v_{1} = \frac{- 1 + 9}{2 \cdot 4}, v_{2} = \frac{- 1 - 9}{2 \cdot 4}

v = \frac{- 1 + 9}{2 \cdot 4} : 1

\frac{- 1 + 9}{2 \cdot 4}

Add/Subtract the numbers:

- 1 + 9 = 8

= \frac{8}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8}{8}

Apply rule

\frac{a}{a} = 1

= 1

v = \frac{- 1 - 9}{2 \cdot 4} : - \frac{5}{4}

\frac{- 1 - 9}{2 \cdot 4}

Subtract the numbers:

- 1 - 9 = - 10

= \frac{- 10}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 10}{8}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{10}{8}

Cancel the common factor:

2

= - \frac{5}{4}

The solutions to the quadratic equation are:

v = 1, v = - \frac{5}{4}

v = 1, v = - \frac{5}{4}

Substitute back

v = 2^{u},

solve for

u

Solve

2^{u} = 1 : u = 0

2^{u} = 1

Apply exponent rules

2^{u} = 1

f (x) = g (x)

, then

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

ln (2^{u}) = ln (1)

Apply log rule:

ln (x^{a}) = a \cdot ln (x)

ln (2^{u}) = u ln (2)

u ln (2) = ln (1)

u ln (2) = ln (1)

Solve

u ln (2) = ln (1) : u = 0

u ln (2) = ln (1)

Divide both sides by

ln (2)

u ln (2) = ln (1)

Divide both sides by

ln (2)

\frac{u ln ( 2 )}{ln ( 2 )} = \frac{ln ( 1 )}{ln ( 2 )}

Simplify

\frac{u ln ( 2 )}{ln ( 2 )} = \frac{ln ( 1 )}{ln ( 2 )}

Simplify

\frac{u ln ( 2 )}{ln ( 2 )} : u

\frac{u ln ( 2 )}{ln ( 2 )}

Cancel the common factor:

ln (2)

= u

Simplify

\frac{ln ( 1 )}{ln ( 2 )} : 0

\frac{ln ( 1 )}{ln ( 2 )}

Simplify

ln (1) : 0

ln (1)

Apply log rule:

lo g_{a} (1) = 0

= 0

= \frac{0}{ln ( 2 )}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = 0

u = 0

u = 0

u = 0

Solve

2^{u} = - \frac{5}{4} :

No Solution for

u \in R

2^{u} = - \frac{5}{4}

a^{f (u)}

cannot be zero or negative for

u \in R

No Solution for u \in R

u = 0

Substitute back

u = arccos (x)

arccos (x) = 0

arccos (x) = 0

arccos (x) = 0 : x = 1

arccos (x) = 0

Apply trig inverse properties

arccos (x) = 0

arccos (x) = a \Rightarrow x = cos (a)

x = cos (0)

cos (0) = 1

cos (0)

Use the following trivial identity:

cos (0) = 1

cos (0)

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 1

= 1

x = 1

x = 1

Combine all the solutions

x = 1

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 2^{arccos(x)}+4^{arccos(x)+1}-2=3 ?
The general solution for 2^{arccos(x)}+4^{arccos(x)+1}-2=3 is x=1