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

The general solution for 2cos^2(x)+3cos(x)-3=0 is x=0.81462…+2pin,x=2pi-0.81462…+2pin

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2cos^2(x)+3cos(x)-3=0

2 cos^{2} (x) + 3 cos (x) - 3 = 0

x = 0.81462 \dots + 2 πn, x = 2 π - 0.81462 \dots + 2 πn

Degrees

x = 46.67462 \dots^{\circ} + 36 0^{\circ} n, x = 313.32537 \dots^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (x) + 3 cos (x) - 3 = 0

Solve by substitution

2 cos^{2} (x) + 3 cos (x) - 3 = 0

Let:

cos (x) = u

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

2 u^{2} + 3 u - 3 = 0 : u = \frac{- 3 + 33}{4}, u = \frac{- 3 - 33}{4}

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

3^{2} - 4 \cdot 2 (- 3) = 33

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 2 \cdot 3 = 24

= 3^{2} + 24

3^{2} = 9

= 9 + 24

Add the numbers:

9 + 24 = 33

= 33

u_{1, 2} = \frac{- 3 \pm 33}{2 \cdot 2}

Separate the solutions

u_{1} = \frac{- 3 + 33}{2 \cdot 2}, u_{2} = \frac{- 3 - 33}{2 \cdot 2}

u = \frac{- 3 + 33}{2 \cdot 2} : \frac{- 3 + 33}{4}

\frac{- 3 + 33}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 3 + 33}{4}

u = \frac{- 3 - 33}{2 \cdot 2} : \frac{- 3 - 33}{4}

\frac{- 3 - 33}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 3 - 33}{4}

The solutions to the quadratic equation are:

u = \frac{- 3 + 33}{4}, u = \frac{- 3 - 33}{4}

Substitute back

u = cos (x)

cos (x) = \frac{- 3 + 33}{4}, cos (x) = \frac{- 3 - 33}{4}

cos (x) = \frac{- 3 + 33}{4}, cos (x) = \frac{- 3 - 33}{4}

cos (x) = \frac{- 3 + 33}{4} : x = arccos (\frac{- 3 + 33}{4}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 33}{4}) + 2 πn

cos (x) = \frac{- 3 + 33}{4}

Apply trig inverse properties

cos (x) = \frac{- 3 + 33}{4}

General solutions for

cos (x) = \frac{- 3 + 33}{4}

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

x = arccos (\frac{- 3 + 33}{4}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 33}{4}) + 2 πn

x = arccos (\frac{- 3 + 33}{4}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 33}{4}) + 2 πn

cos (x) = \frac{- 3 - 33}{4} :

No Solution

cos (x) = \frac{- 3 - 33}{4}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = arccos (\frac{- 3 + 33}{4}) + 2 πn, x = 2 π - arccos (\frac{- 3 + 33}{4}) + 2 πn

Show solutions in decimal form

x = 0.81462 \dots + 2 πn, x = 2 π - 0.81462 \dots + 2 πn

3 sin (x) - 5 cos (x) = 0

sin^{2} (x) = 0.85

cos (x - 6 0^{\circ}) = \frac{3}{2}

(0.5 \cdot 9.8 \cdot l (1 - cos (θ))) = 0.018

x, 0 = - sin (2 x) - 2 cos (x)

What is the general solution for 2cos^2(x)+3cos(x)-3=0 ?
The general solution for 2cos^2(x)+3cos(x)-3=0 is x=0.81462…+2pin,x=2pi-0.81462…+2pin