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Popular >

2cos^2(θ)-sqrt(3)cos(θ)=0,0<= θ<2pi

Solution

2 cos^{2} (θ) - 3 cos (θ) = 0, 0 \leq θ < 2 π

Solution

θ = \frac{π}{6}, θ = \frac{11 π}{6}, θ = \frac{π}{2}, θ = \frac{3 π}{2}

Degrees

θ = 3 0^{\circ}, θ = 33 0^{\circ}, θ = 9 0^{\circ}, θ = 27 0^{\circ}

Solution steps

2 cos^{2} (θ) - 3 cos (θ) = 0, 0 \leq θ < 2 π

Solve by substitution

2 cos^{2} (θ) - 3 cos (θ) = 0

Let:

cos (θ) = u

2 u^{2} - 3 u = 0

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

2 u^{2} - 3 u = 0

Solve with the quadratic formula

2 u^{2} - 3 u = 0

Quadratic Equation Formula:

For

a = 2, b = - 3, c = 0

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

u_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 2 \cdot 0}{2 \cdot 2}

(- 3)^{2} - 4 \cdot 2 \cdot 0 = 3

(- 3)^{2} - 4 \cdot 2 \cdot 0

(- 3)^{2} = 3

(- 3)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 3)^{2} = (3)^{2}

= (3)^{2}

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

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

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 3 - 0

Subtract the numbers:

3 - 0 = 3

= 3

u_{1, 2} = \frac{- ( - 3 ) \pm 3}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 3}{2 \cdot 2}, u_{2} = \frac{- ( - 3 ) - 3}{2 \cdot 2}

u = \frac{- ( - 3 ) + 3}{2 \cdot 2} : \frac{3}{2}

\frac{- ( - 3 ) + 3}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{3 + 3}{2 \cdot 2}

Add similar elements:

3 + 3 = 23

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Cancel the common factor:

2

= \frac{3}{2}

u = \frac{- ( - 3 ) - 3}{2 \cdot 2} : 0

\frac{- ( - 3 ) - 3}{2 \cdot 2}

Apply rule

- (- a) = a

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

Add similar elements:

3 - 3 = 0

= \frac{0}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

The solutions to the quadratic equation are:

u = \frac{3}{2}, u = 0

Substitute back

u = cos (θ)

cos (θ) = \frac{3}{2}, cos (θ) = 0

cos (θ) = \frac{3}{2}, cos (θ) = 0

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

cos (θ) = \frac{3}{2}, 0 \leq θ < 2 π

General solutions for

cos (θ) = \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}

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

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{6}, θ = \frac{11 π}{6}

cos (θ) = 0, 0 \leq θ < 2 π : θ = \frac{π}{2}, θ = \frac{3 π}{2}

cos (θ) = 0, 0 \leq θ < 2 π

General solutions for

cos (θ) = 0

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}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

Combine all the solutions

θ = \frac{π}{6}, θ = \frac{11 π}{6}, θ = \frac{π}{2}, θ = \frac{3 π}{2}

Popular Examples

tan(θ)=(-1)/3

tan (θ) = \frac{- 1}{3}

tan(θ)=(-1)/2

tan (θ) = \frac{- 1}{2}

sin(α)= 1/2

sin (α) = \frac{1}{2}

sqrt(3)sin(3x)+3cos(3x)=0

3 sin (3 x) + 3 cos (3 x) = 0

tan(θ)=-4/5 , pi/2 <θ<pi

tan (θ) = - \frac{4}{5}, \frac{π}{2} < θ < π