What is the general solution for sin^2(θ)+2cos(θ)= 7/4 ?

The general solution for sin^2(θ)+2cos(θ)= 7/4 is θ= pi/3+2pin,θ=(5pi)/3+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin^2(θ)+2cos(θ)= 7/4

Solution

sin^{2} (θ) + 2 cos (θ) = \frac{7}{4}

Solution

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

Degrees

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

Solution steps

sin^{2} (θ) + 2 cos (θ) = \frac{7}{4}

Subtract

\frac{7}{4}

from both sides

sin^{2} (θ) + 2 cos (θ) - \frac{7}{4} = 0

Simplify

sin^{2} (θ) + 2 cos (θ) - \frac{7}{4} : \frac{4 sin ^{2} ( θ ) + 8 cos ( θ ) - 7}{4}

sin^{2} (θ) + 2 cos (θ) - \frac{7}{4}

Convert element to fraction:

sin^{2} (θ) = \frac{sin ^{2} ( θ ) 4}{4}, 2 cos (θ) = \frac{2 cos ( θ ) 4}{4}

= \frac{sin ^{2} ( θ ) \cdot 4}{4} + \frac{2 cos ( θ ) \cdot 4}{4} - \frac{7}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ^{2} ( θ ) \cdot 4 + 2 cos ( θ ) \cdot 4 - 7}{4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{4 sin ^{2} ( θ ) + 8 cos ( θ ) - 7}{4}

\frac{4 sin ^{2} ( θ ) + 8 cos ( θ ) - 7}{4} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 sin^{2} (θ) + 8 cos (θ) - 7 = 0

Rewrite using trig identities

- 7 + 4 sin^{2} (θ) + 8 cos (θ)

Use the Pythagorean identity:

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

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

= - 7 + 4 (1 - cos^{2} (θ)) + 8 cos (θ)

Simplify

- 7 + 4 (1 - cos^{2} (θ)) + 8 cos (θ) : 8 cos (θ) - 4 cos^{2} (θ) - 3

- 7 + 4 (1 - cos^{2} (θ)) + 8 cos (θ)

Expand

4 (1 - cos^{2} (θ)) : 4 - 4 cos^{2} (θ)

4 (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = cos^{2} (θ)

= 4 \cdot 1 - 4 cos^{2} (θ)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 cos^{2} (θ)

= - 7 + 4 - 4 cos^{2} (θ) + 8 cos (θ)

Add/Subtract the numbers:

- 7 + 4 = - 3

= 8 cos (θ) - 4 cos^{2} (θ) - 3

= 8 cos (θ) - 4 cos^{2} (θ) - 3

- 3 - 4 cos^{2} (θ) + 8 cos (θ) = 0

Solve by substitution

- 3 - 4 cos^{2} (θ) + 8 cos (θ) = 0

Let:

cos (θ) = u

- 3 - 4 u^{2} + 8 u = 0

- 3 - 4 u^{2} + 8 u = 0 : u = \frac{1}{2}, u = \frac{3}{2}

- 3 - 4 u^{2} + 8 u = 0

Write in the standard form

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

- 4 u^{2} + 8 u - 3 = 0

Solve with the quadratic formula

- 4 u^{2} + 8 u - 3 = 0

Quadratic Equation Formula:

For

a = - 4, b = 8, c = - 3

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

u_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) ( - 3 )}{2 ( - 4 )}

8^{2} - 4 (- 4) (- 3) = 4

8^{2} - 4 (- 4) (- 3)

Apply rule

- (- a) = a

= 8^{2} - 4 \cdot 4 \cdot 3

Multiply the numbers:

4 \cdot 4 \cdot 3 = 48

= 8^{2} - 48

8^{2} = 64

= 64 - 48

Subtract the numbers:

64 - 48 = 16

= 16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

u_{1, 2} = \frac{- 8 \pm 4}{2 ( - 4 )}

Separate the solutions

u_{1} = \frac{- 8 + 4}{2 ( - 4 )}, u_{2} = \frac{- 8 - 4}{2 ( - 4 )}

u = \frac{- 8 + 4}{2 ( - 4 )} : \frac{1}{2}

\frac{- 8 + 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 8 + 4}{- 2 \cdot 4}

Add/Subtract the numbers:

- 8 + 4 = - 4

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

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 4}{- 8}

Apply the fraction rule:

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

= \frac{4}{8}

Cancel the common factor:

4

= \frac{1}{2}

u = \frac{- 8 - 4}{2 ( - 4 )} : \frac{3}{2}

\frac{- 8 - 4}{2 ( - 4 )}

Remove parentheses:

(- a) = - a

= \frac{- 8 - 4}{- 2 \cdot 4}

Subtract the numbers:

- 8 - 4 = - 12

= \frac{- 12}{- 2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{- 12}{- 8}

Apply the fraction rule:

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

= \frac{12}{8}

Cancel the common factor:

4

= \frac{3}{2}

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = \frac{3}{2}

Substitute back

u = cos (θ)

cos (θ) = \frac{1}{2}, cos (θ) = \frac{3}{2}

cos (θ) = \frac{1}{2}, cos (θ) = \frac{3}{2}

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

cos (θ) = \frac{1}{2}

General solutions for

cos (θ) = \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}

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

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

cos (θ) = \frac{3}{2} :

No Solution

cos (θ) = \frac{3}{2}

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

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

Popular Examples

5cos(x)+3=3cos(x)+5

5 cos (x) + 3 = 3 cos (x) + 5

sin(pi/2)=cos(x)

sin (\frac{π}{2}) = cos (x)

3sin^2(x)=3-sin^2(x)

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

solvefor x,arcsin(x/(16))=31 solve for

x, arcsin (\frac{x}{16}) = 31

sin(x)(2cos^2(x)-1)=0

sin (x) (2 cos^{2} (x) - 1) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)+2cos(θ)= 7/4 ?
The general solution for sin^2(θ)+2cos(θ)= 7/4 is θ= pi/3+2pin,θ=(5pi)/3+2pin