What is the general solution for 2sin^2(u)-3cos(u)=0 ?

The general solution for 2sin^2(u)-3cos(u)=0 is u= pi/3+2pin,u=(5pi)/3+2pin

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2sin^2(u)-3cos(u)=0

Solution

2 sin^{2} (u) - 3 cos (u) = 0

Solution

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

Degrees

u = 6 0^{\circ} + 36 0^{\circ} n, u = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin^{2} (u) - 3 cos (u) = 0

Rewrite using trig identities

2 sin^{2} (u) - 3 cos (u)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

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

= 2 (1 - cos^{2} (u)) - 3 cos (u)

(1 - cos^{2} (u)) \cdot 2 - 3 cos (u) = 0

Solve by substitution

(1 - cos^{2} (u)) \cdot 2 - 3 cos (u) = 0

Let:

cos (u) = v

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

(1 - v^{2}) \cdot 2 - 3 v = 0 : v = - 2, v = \frac{1}{2}

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

Expand

(1 - v^{2}) \cdot 2 - 3 v : 2 - 2 v^{2} - 3 v

(1 - v^{2}) \cdot 2 - 3 v

= 2 (1 - v^{2}) - 3 v

Expand

2 (1 - v^{2}) : 2 - 2 v^{2}

2 (1 - v^{2})

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = v^{2}

= 2 \cdot 1 - 2 v^{2}

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 v^{2}

= 2 - 2 v^{2} - 3 v

2 - 2 v^{2} - 3 v = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 v^{2} - 3 v + 2 = 0

Solve with the quadratic formula

- 2 v^{2} - 3 v + 2 = 0

Quadratic Equation Formula:

For

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

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

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

(- 3)^{2} - 4 (- 2) \cdot 2 = 5

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 3^{2} + 4 \cdot 2 \cdot 2

Multiply the numbers:

4 \cdot 2 \cdot 2 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

v_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 ( - 2 )}

Separate the solutions

v_{1} = \frac{- ( - 3 ) + 5}{2 ( - 2 )}, v_{2} = \frac{- ( - 3 ) - 5}{2 ( - 2 )}

v = \frac{- ( - 3 ) + 5}{2 ( - 2 )} : - 2

\frac{- ( - 3 ) + 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 + 5}{- 2 \cdot 2}

Add the numbers:

3 + 5 = 8

= \frac{8}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{8}{- 4}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{8}{4}

Divide the numbers:

\frac{8}{4} = 2

= - 2

v = \frac{- ( - 3 ) - 5}{2 ( - 2 )} : \frac{1}{2}

\frac{- ( - 3 ) - 5}{2 ( - 2 )}

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{3 - 5}{- 2 \cdot 2}

Subtract the numbers:

3 - 5 = - 2

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 2}{- 4}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

The solutions to the quadratic equation are:

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

Substitute back

v = cos (u)

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

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

cos (u) = - 2 :

No Solution

cos (u) = - 2

- 1 \leq cos (x) \leq 1

No Solution

cos (u) = \frac{1}{2} : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

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

General solutions for

cos (u) = \frac{1}{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}

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

Combine all the solutions

u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

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cos (x) = \frac{17.6}{26}

tan(2θ)=-2.5

tan (2 θ) = - 2.5

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\frac{150}{sin ( x )} = \frac{140.78}{sin ( 11 5 ^{\circ} )}

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(u)-3cos(u)=0 ?
The general solution for 2sin^2(u)-3cos(u)=0 is u= pi/3+2pin,u=(5pi)/3+2pin