What is the general solution for cos(2x)=1-2(3/5)^2 ?

The general solution for cos(2x)=1-2(3/5)^2 is x=(1.28700…)/2+pin,x=pi-(1.28700…)/2+pin

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cos(2x)=1-2(3/5)^2

cos (2 x) = 1 - 2 (\frac{3}{5})^{2}

x = \frac{1.28700 \dots}{2} + πn, x = π - \frac{1.28700 \dots}{2} + πn

Radians

x = 0.64350 \dots + πn, x = π - 0.64350 \dots + πn

Solution steps

cos (2 x) = 1 - 2 (\frac{3}{5})^{2}

Simplify

1 - 2 (\frac{3}{5})^{2} : \frac{7}{25}

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

2 (\frac{3}{5})^{2} = \frac{18}{25}

2 (\frac{3}{5})^{2}

(\frac{3}{5})^{2} = \frac{3 ^{2}}{5 ^{2}}

(\frac{3}{5})^{2}

Apply exponent rule:

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

= \frac{3 ^{2}}{5 ^{2}}

= 2 \cdot \frac{3 ^{2}}{5 ^{2}}

Multiply fractions:

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

= \frac{3 ^{2} \cdot 2}{5 ^{2}}

3^{2} \cdot 2 = 18

3^{2} \cdot 2

3^{2} = 9

= 9 \cdot 2

Multiply the numbers:

9 \cdot 2 = 18

= 18

= \frac{18}{5 ^{2}}

5^{2} = 25

= \frac{18}{25}

= 1 - \frac{18}{25}

Convert element to fraction:

1 = \frac{1 \cdot 25}{25}

= \frac{1 \cdot 25}{25} - \frac{18}{25}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 25 - 18}{25}

1 \cdot 25 - 18 = 7

1 \cdot 25 - 18

Multiply the numbers:

1 \cdot 25 = 25

= 25 - 18

Subtract the numbers:

25 - 18 = 7

= 7

= \frac{7}{25}

Apply trig inverse properties

cos (2 x) = \frac{7}{25}

General solutions for

cos (2 x) = \frac{7}{25}

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

2 x = arccos (\frac{7}{25}) + 2 πn, 2 x = 2 π - arccos (\frac{7}{25}) + 2 πn

2 x = arccos (\frac{7}{25}) + 2 πn, 2 x = 2 π - arccos (\frac{7}{25}) + 2 πn

Solve

2 x = arccos (\frac{7}{25}) + 2 πn : x = \frac{arccos ( \frac{7}{25} )}{2} + πn

2 x = arccos (\frac{7}{25}) + 2 πn

Divide both sides by

2

2 x = arccos (\frac{7}{25}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{arccos ( \frac{7}{25} )}{2} + \frac{2 πn}{2}

Simplify

x = \frac{arccos ( \frac{7}{25} )}{2} + πn

x = \frac{arccos ( \frac{7}{25} )}{2} + πn

Solve

2 x = 2 π - arccos (\frac{7}{25}) + 2 πn : x = π - \frac{arccos ( \frac{7}{25} )}{2} + πn

2 x = 2 π - arccos (\frac{7}{25}) + 2 πn

Divide both sides by

2

2 x = 2 π - arccos (\frac{7}{25}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 π}{2} - \frac{arccos ( \frac{7}{25} )}{2} + \frac{2 πn}{2}

Simplify

x = π - \frac{arccos ( \frac{7}{25} )}{2} + πn

x = π - \frac{arccos ( \frac{7}{25} )}{2} + πn

x = \frac{arccos ( \frac{7}{25} )}{2} + πn, x = π - \frac{arccos ( \frac{7}{25} )}{2} + πn

Show solutions in decimal form

x = \frac{1.28700 \dots}{2} + πn, x = π - \frac{1.28700 \dots}{2} + πn

sin (θ + \frac{π}{3}) = \frac{1}{2}

4 sin^{2} (x) = 5 - 4 cos (x), 0 \leq x < 2 π

sin (x) - 6 cos^{2} (x) + 4 = 0

cos (θ) = \frac{3}{4}

2 sec (4 x) - 2 = 0

What is the general solution for cos(2x)=1-2(3/5)^2 ?
The general solution for cos(2x)=1-2(3/5)^2 is x=(1.28700…)/2+pin,x=pi-(1.28700…)/2+pin