What is the general solution for arccos(x)=2arccos(3/4) ?

The general solution for arccos(x)=2arccos(3/4) is x= 1/8

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arccos(x)=2arccos(3/4)

arccos (x) = 2 arccos (\frac{3}{4})

x = \frac{1}{8}

Solution steps

arccos (x) = 2 arccos (\frac{3}{4})

Apply trig inverse properties

arccos (x) = 2 arccos (\frac{3}{4})

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

x = cos (2 arccos (\frac{3}{4}))

cos (2 arccos (\frac{3}{4})) = \frac{1}{8}

cos (2 arccos (\frac{3}{4}))

Rewrite using trig identities:

2 cos^{2} (arccos (\frac{3}{4})) - 1

cos (2 arccos (\frac{3}{4}))

Use the Double Angle identity:

cos (2 x) = 2 cos^{2} (x) - 1

= 2 cos^{2} (arccos (\frac{3}{4})) - 1

= 2 cos^{2} (arccos (\frac{3}{4})) - 1

Rewrite using trig identities:

cos (arccos (\frac{3}{4})) = \frac{3}{4}

Use the following identity:

cos (arccos (x)) = x

= \frac{3}{4}

= 2 (\frac{3}{4})^{2} - 1

Simplify

2 (\frac{3}{4})^{2} - 1 : \frac{1}{8}

2 (\frac{3}{4})^{2} - 1

2 (\frac{3}{4})^{2} = \frac{9}{8}

2 (\frac{3}{4})^{2}

(\frac{3}{4})^{2} = \frac{3 ^{2}}{4 ^{2}}

(\frac{3}{4})^{2}

Apply exponent rule:

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

= \frac{3 ^{2}}{4 ^{2}}

= 2 \cdot \frac{3 ^{2}}{4 ^{2}}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 ^{2} \cdot 2}{4 ^{2}}

Factor

4^{2} : 2^{4}

Factor

4 = 2^{2}

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

Simplify

(2^{2})^{2} : 2^{4}

(2^{2})^{2}

Apply exponent rule:

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

= 2^{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 2^{4}

= 2^{4}

= \frac{3 ^{2} \cdot 2}{2 ^{4}}

Cancel the common factor:

2

= \frac{3 ^{2}}{2 ^{3}}

3^{2} = 9

= \frac{9}{2 ^{3}}

2^{3} = 8

= \frac{9}{8}

= \frac{9}{8} - 1

Convert element to fraction:

1 = \frac{1 \cdot 8}{8}

= - \frac{1 \cdot 8}{8} + \frac{9}{8}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 \cdot 8 + 9}{8}

- 1 \cdot 8 + 9 = 1

- 1 \cdot 8 + 9

Multiply the numbers:

1 \cdot 8 = 8

= - 8 + 9

Add/Subtract the numbers:

- 8 + 9 = 1

= 1

= \frac{1}{8}

= \frac{1}{8}

x = \frac{1}{8}

x = \frac{1}{8}

2 sin (x) - 3 = 0

2 cos^{2} (x) - 3 cos (x) + 1 = 0

sin (x) = cos (2 x)

cos^{3} (x) = cos (x)

2 cos (x) = 1

What is the general solution for arccos(x)=2arccos(3/4) ?
The general solution for arccos(x)=2arccos(3/4) is x= 1/8