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

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

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

2 cos^{2} (x) = 1, 0 \leq x \leq π

x = \frac{π}{4}, x = \frac{3 π}{4}

Degrees

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

Solution steps

2 cos^{2} (x) = 1, 0 \leq x \leq π

Solve by substitution

2 cos^{2} (x) = 1

Let:

cos (x) = u

2 u^{2} = 1

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

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 \leq π : x = \frac{π}{4}

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

x = \frac{π}{4}

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

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

x = \frac{3 π}{4}

Combine all the solutions

x = \frac{π}{4}, x = \frac{3 π}{4}

3 sin (x) = 2 tan (x)

2 cos (x) = sin (2 x)

cos (θ) = 12

cos^{2} (x) + 2 cos (x) = 0

2 sin (x) = tan (2 x)

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