What is the general solution for 2arctan(1/2)=arccos(x) ?

The general solution for 2arctan(1/2)=arccos(x) is x= 3/5

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2arctan(1/2)=arccos(x)

2 arctan (\frac{1}{2}) = arccos (x)

x = \frac{3}{5}

Solution steps

2 arctan (\frac{1}{2}) = arccos (x)

Switch sides

arccos (x) = 2 arctan (\frac{1}{2})

Apply trig inverse properties

arccos (x) = 2 arctan (\frac{1}{2})

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

x = cos (2 arctan (\frac{1}{2}))

cos (2 arctan (\frac{1}{2})) = \frac{3}{5}

cos (2 arctan (\frac{1}{2}))

Rewrite using trig identities:

1 - 2 sin^{2} (arctan (\frac{1}{2}))

cos (2 arctan (\frac{1}{2}))

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (arctan (\frac{1}{2}))

= 1 - 2 sin^{2} (arctan (\frac{1}{2}))

Rewrite using trig identities:

sin (arctan (\frac{1}{2})) = \frac{5}{5}

sin (arctan (\frac{1}{2}))

Rewrite using trig identities:

sin (arctan (\frac{1}{2})) = \frac{( \frac{1}{2} ) 1 + ( \frac{1}{2} ) ^{2}}{1 + ( \frac{1}{2} ) ^{2}}

Use the following identity:

sin (arctan (x)) = \frac{x 1 + x ^{2}}{1 + x ^{2}}

= \frac{( \frac{1}{2} ) 1 + ( \frac{1}{2} ) ^{2}}{1 + ( \frac{1}{2} ) ^{2}}

= \frac{\frac{1}{2} 1 + ( \frac{1}{2} ) ^{2}}{1 + ( \frac{1}{2} ) ^{2}}

Simplify

= \frac{5}{5}

= 1 - 2 (\frac{5}{5})^{2}

Simplify

1 - 2 (\frac{5}{5})^{2} : \frac{3}{5}

1 - 2 (\frac{5}{5})^{2}

2 (\frac{5}{5})^{2} = \frac{2}{5}

2 (\frac{5}{5})^{2}

(\frac{5}{5})^{2} = \frac{1}{5}

(\frac{5}{5})^{2}

Apply exponent rule:

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

= \frac{( 5 ) ^{2}}{5 ^{2}}

(5)^{2} : 5

Apply radical rule:

a = a^{\frac{1}{2}}

= (5^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 5^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 5

= \frac{5}{5 ^{2}}

Cancel the common factor:

5

= \frac{1}{5}

= 2 \cdot \frac{1}{5}

Multiply fractions:

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

= \frac{1 \cdot 2}{5}

Multiply the numbers:

1 \cdot 2 = 2

= \frac{2}{5}

= 1 - \frac{2}{5}

Convert element to fraction:

1 = \frac{1 \cdot 5}{5}

= \frac{1 \cdot 5}{5} - \frac{2}{5}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 5 - 2}{5}

1 \cdot 5 - 2 = 3

1 \cdot 5 - 2

Multiply the numbers:

1 \cdot 5 = 5

= 5 - 2

Subtract the numbers:

5 - 2 = 3

= 3

= \frac{3}{5}

= \frac{3}{5}

x = \frac{3}{5}

x = \frac{3}{5}

2 sin^{2} (x) - 7 sin (x) = - 3

sin (\frac{x}{2}) + sin (\frac{x}{2}) = sin (x)

cos (x) + 2 sin^{2} (x) = 1

15 cos (x) - 5 = 4 cos (x)

sin (θ) + cos (θ) = \frac{2 + 3}{2}

What is the general solution for 2arctan(1/2)=arccos(x) ?
The general solution for 2arctan(1/2)=arccos(x) is x= 3/5