What is the general solution for cos(u)-1.5sin^2(u)+0.1667=0 ?

The general solution for cos(u)-1.5sin^2(u)+0.1667=0 is u=0.84108…+2pin,u=2pi-0.84108…+2pin

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cos(u)-1.5sin^2(u)+0.1667=0

Solution

cos (u) - 1.5 sin^{2} (u) + 0.1667 = 0

Solution

u = 0.84108 \dots + 2 πn, u = 2 π - 0.84108 \dots + 2 πn

Degrees

u = 48.19053 \dots^{\circ} + 36 0^{\circ} n, u = 311.80946 \dots^{\circ} + 36 0^{\circ} n

Solution steps

cos (u) - 1.5 sin^{2} (u) + 0.1667 = 0

Rewrite using trig identities

0.1667 + cos (u) - 1.5 sin^{2} (u)

Use the Pythagorean identity:

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

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

= 0.1667 + cos (u) - 1.5 (1 - cos^{2} (u))

Simplify

0.1667 + cos (u) - 1.5 (1 - cos^{2} (u)) : 1.5 cos^{2} (u) + cos (u) - 1.3333

0.1667 + cos (u) - 1.5 (1 - cos^{2} (u))

Expand

- 1.5 (1 - cos^{2} (u)) : - 1.5 + 1.5 cos^{2} (u)

- 1.5 (1 - cos^{2} (u))

Apply the distributive law:

a (b - c) = ab - a c

a = - 1.5, b = 1, c = cos^{2} (u)

= - 1.5 \cdot 1 - (- 1.5) cos^{2} (u)

Apply minus-plus rules

- (- a) = a

= - 1 \cdot 1.5 + 1.5 cos^{2} (u)

Multiply the numbers:

1 \cdot 1.5 = 1.5

= - 1.5 + 1.5 cos^{2} (u)

= 0.1667 + cos (u) - 1.5 + 1.5 cos^{2} (u)

Simplify

0.1667 + cos (u) - 1.5 + 1.5 cos^{2} (u) : 1.5 cos^{2} (u) + cos (u) - 1.3333

0.1667 + cos (u) - 1.5 + 1.5 cos^{2} (u)

Group like terms

= cos (u) + 1.5 cos^{2} (u) + 0.1667 - 1.5

Add/Subtract the numbers:

0.1667 - 1.5 = - 1.3333

= 1.5 cos^{2} (u) + cos (u) - 1.3333

= 1.5 cos^{2} (u) + cos (u) - 1.3333

= 1.5 cos^{2} (u) + cos (u) - 1.3333

- 1.3333 + cos (u) + 1.5 cos^{2} (u) = 0

Solve by substitution

- 1.3333 + cos (u) + 1.5 cos^{2} (u) = 0

Let:

cos (u) = v

- 1.3333 + v + 1.5 v^{2} = 0

- 1.3333 + v + 1.5 v^{2} = 0 : v = \frac{- 1 + 8.9998}{3}, v = \frac{- 1 - 8.9998}{3}

- 1.3333 + v + 1.5 v^{2} = 0

Write in the standard form

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

1.5 v^{2} + v - 1.3333 = 0

Solve with the quadratic formula

1.5 v^{2} + v - 1.3333 = 0

Quadratic Equation Formula:

For

a = 1.5, b = 1, c = - 1.3333

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1.5 ( - 1.3333 )}{2 \cdot 1.5}

v_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1.5 ( - 1.3333 )}{2 \cdot 1.5}

1^{2} - 4 \cdot 1.5 (- 1.3333) = 8.9998

1^{2} - 4 \cdot 1.5 (- 1.3333)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1.3333) \cdot 1.5

Apply rule

- (- a) = a

= 1 + 4 \cdot 1.5 \cdot 1.3333

Multiply the numbers:

4 \cdot 1.5 \cdot 1.3333 = 7.9998

= 1 + 7.9998

Add the numbers:

1 + 7.9998 = 8.9998

= 8.9998

v_{1, 2} = \frac{- 1 \pm 8.9998}{2 \cdot 1.5}

Separate the solutions

v_{1} = \frac{- 1 + 8.9998}{2 \cdot 1.5}, v_{2} = \frac{- 1 - 8.9998}{2 \cdot 1.5}

v = \frac{- 1 + 8.9998}{2 \cdot 1.5} : \frac{- 1 + 8.9998}{3}

\frac{- 1 + 8.9998}{2 \cdot 1.5}

Multiply the numbers:

2 \cdot 1.5 = 3

= \frac{- 1 + 8.9998}{3}

v = \frac{- 1 - 8.9998}{2 \cdot 1.5} : \frac{- 1 - 8.9998}{3}

\frac{- 1 - 8.9998}{2 \cdot 1.5}

Multiply the numbers:

2 \cdot 1.5 = 3

= \frac{- 1 - 8.9998}{3}

The solutions to the quadratic equation are:

v = \frac{- 1 + 8.9998}{3}, v = \frac{- 1 - 8.9998}{3}

Substitute back

v = cos (u)

cos (u) = \frac{- 1 + 8.9998}{3}, cos (u) = \frac{- 1 - 8.9998}{3}

cos (u) = \frac{- 1 + 8.9998}{3}, cos (u) = \frac{- 1 - 8.9998}{3}

cos (u) = \frac{- 1 + 8.9998}{3} : u = arccos (\frac{- 1 + 8.9998}{3}) + 2 πn, u = 2 π - arccos (\frac{- 1 + 8.9998}{3}) + 2 πn

cos (u) = \frac{- 1 + 8.9998}{3}

Apply trig inverse properties

cos (u) = \frac{- 1 + 8.9998}{3}

General solutions for

cos (u) = \frac{- 1 + 8.9998}{3}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

u = arccos (\frac{- 1 + 8.9998}{3}) + 2 πn, u = 2 π - arccos (\frac{- 1 + 8.9998}{3}) + 2 πn

u = arccos (\frac{- 1 + 8.9998}{3}) + 2 πn, u = 2 π - arccos (\frac{- 1 + 8.9998}{3}) + 2 πn

cos (u) = \frac{- 1 - 8.9998}{3} :

No Solution

cos (u) = \frac{- 1 - 8.9998}{3}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

u = arccos (\frac{- 1 + 8.9998}{3}) + 2 πn, u = 2 π - arccos (\frac{- 1 + 8.9998}{3}) + 2 πn

Show solutions in decimal form

u = 0.84108 \dots + 2 πn, u = 2 π - 0.84108 \dots + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for cos(u)-1.5sin^2(u)+0.1667=0 ?
The general solution for cos(u)-1.5sin^2(u)+0.1667=0 is u=0.84108…+2pin,u=2pi-0.84108…+2pin