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

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

The general solution for cos^2(x)-3/4 =0,0<= x<= 2pi is x= pi/6 ,x=(11pi)/6 ,x=(5pi)/6 ,x=(7pi)/6

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cos^2(x)-3/4 =0,0<= x<= 2pi

Solution

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

Solution

x = \frac{π}{6}, x = \frac{11 π}{6}, x = \frac{5 π}{6}, x = \frac{7 π}{6}

Degrees

x = 3 0^{\circ}, x = 33 0^{\circ}, x = 15 0^{\circ}, x = 21 0^{\circ}

Solution steps

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

Solve by substitution

cos^{2} (x) - \frac{3}{4} = 0

Let:

cos (x) = u

u^{2} - \frac{3}{4} = 0

u^{2} - \frac{3}{4} = 0 : u = \frac{3}{2}, u = - \frac{3}{2}

u^{2} - \frac{3}{4} = 0

Move

\frac{3}{4}

to the right side

u^{2} - \frac{3}{4} = 0

Add

\frac{3}{4}

to both sides

u^{2} - \frac{3}{4} + \frac{3}{4} = 0 + \frac{3}{4}

Simplify

u^{2} = \frac{3}{4}

u^{2} = \frac{3}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{3}{4}, u = - \frac{3}{4}

\frac{3}{4} = \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{4} = - \frac{3}{2}

- \frac{3}{4}

Simplify

\frac{3}{4} : \frac{3}{2}

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

= - \frac{3}{2}

u = \frac{3}{2}, u = - \frac{3}{2}

Substitute back

u = cos (x)

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{3}{2}, cos (x) = - \frac{3}{2}

cos (x) = \frac{3}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{11 π}{6}

cos (x) = \frac{3}{2}, 0 \leq x \leq 2 π

General solutions for

cos (x) = \frac{3}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{11 π}{6}

cos (x) = - \frac{3}{2}, 0 \leq x \leq 2 π : x = \frac{5 π}{6}, x = \frac{7 π}{6}

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

General solutions for

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

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{5 π}{6}, x = \frac{7 π}{6}

Combine all the solutions

x = \frac{π}{6}, x = \frac{11 π}{6}, x = \frac{5 π}{6}, x = \frac{7 π}{6}

Popular Examples

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12cos^2(x)+sin(x)-6=0

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

sin(A)= 3/5

sin (A) = \frac{3}{5}

tan(θ)=0.2

tan (θ) = 0.2

sin(x)tan(x)=sin(x),0<= x<= 2pi

sin (x) tan (x) = sin (x), 0 \leq x \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(x)-3/4 =0,0<= x<= 2pi ?
The general solution for cos^2(x)-3/4 =0,0<= x<= 2pi is x= pi/6 ,x=(11pi)/6 ,x=(5pi)/6 ,x=(7pi)/6