What is the general solution for 2cos^2(x)-1=0,0<= x<2pi ?

Q: What is the general solution for 2cos^2(x)-1=0,0<= x<2pi ?

The general solution for 2cos^2(x)-1=0,0<= x<2pi is x= pi/4 ,x=(7pi)/4 ,x=(3pi)/4 ,x=(5pi)/4

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2cos^2(x)-1=0,0<= x<2pi

Solution

2 cos^{2} (x) - 1 = 0, 0 \leq x < 2 π

Solution

x = \frac{π}{4}, x = \frac{7 π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}

Degrees

x = 4 5^{\circ}, x = 31 5^{\circ}, x = 13 5^{\circ}, x = 22 5^{\circ}

Solution steps

2 cos^{2} (x) - 1 = 0, 0 \leq x < 2 π

Solve by substitution

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

Let:

cos (x) = u

2 u^{2} - 1 = 0

2 u^{2} - 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

2 u^{2} - 1 = 0

Move

1

to the right side

2 u^{2} - 1 = 0

Add

1

to both sides

2 u^{2} - 1 + 1 = 0 + 1

Simplify

2 u^{2} = 1

2 u^{2} = 1

Divide both sides by

2

2 u^{2} = 1

Divide both sides by

2

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = cos (x)

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2}

cos (x) = \frac{1}{2}, cos (x) = - \frac{1}{2}

cos (x) = \frac{1}{2}, 0 \leq x < 2 π : x = \frac{π}{4}, x = \frac{7 π}{4}

cos (x) = \frac{1}{2}, 0 \leq x < 2 π

Apply trig inverse properties

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

General solutions for

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

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

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

x = arccos (\frac{1}{2}) + 2 πn, x = 2 π - arccos (\frac{1}{2}) + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{4}, x = \frac{7 π}{4}

cos (x) = - \frac{1}{2}, 0 \leq x < 2 π : x = \frac{3 π}{4}, x = \frac{5 π}{4}

cos (x) = - \frac{1}{2}, 0 \leq x < 2 π

Apply trig inverse properties

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

General solutions for

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

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

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

x = arccos (- \frac{1}{2}) + 2 πn, x = - arccos (- \frac{1}{2}) + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{3 π}{4}, x = \frac{5 π}{4}

Combine all the solutions

x = \frac{π}{4}, x = \frac{7 π}{4}, x = \frac{3 π}{4}, x = \frac{5 π}{4}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(x)-1=0,0<= x<2pi ?
The general solution for 2cos^2(x)-1=0,0<= x<2pi is x= pi/4 ,x=(7pi)/4 ,x=(3pi)/4 ,x=(5pi)/4