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

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

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

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

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

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

Move

\frac{1}{2}

to the right side

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

Add

\frac{1}{2}

to both sides

u^{2} - \frac{1}{2} + \frac{1}{2} = 0 + \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 2 π : x = \frac{π}{4}, x = \frac{7 π}{4}

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

x = \frac{π}{4}, x = \frac{7 π}{4}

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

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

x, tan (x) = tan (y)

4 cos (θ) + 3 = 0

2 cos^{2} (x) + 1 = 3 cos (x), 0 \leq x \leq 2 π

cos (x) = \frac{5}{9}

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

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