What is the general solution for sin(x)+cos^2(x)= 5/4 ?

The general solution for sin(x)+cos^2(x)= 5/4 is x= pi/6+2pin,x=(5pi)/6+2pin

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sin(x)+cos^2(x)= 5/4

Solution

sin (x) + cos^{2} (x) = \frac{5}{4}

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) + cos^{2} (x) = \frac{5}{4}

Subtract

\frac{5}{4}

from both sides

sin (x) + cos^{2} (x) - \frac{5}{4} = 0

Simplify

sin (x) + cos^{2} (x) - \frac{5}{4} : \frac{4 sin ( x ) + 4 cos ^{2} ( x ) - 5}{4}

sin (x) + cos^{2} (x) - \frac{5}{4}

Convert element to fraction:

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

= \frac{sin ( x ) \cdot 4}{4} + \frac{cos ^{2} ( x ) \cdot 4}{4} - \frac{5}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( x ) \cdot 4 + cos ^{2} ( x ) \cdot 4 - 5}{4}

\frac{4 sin ( x ) + 4 cos ^{2} ( x ) - 5}{4} = 0

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

4 sin (x) + 4 cos^{2} (x) - 5 = 0

Rewrite using trig identities

- 5 + 4 cos^{2} (x) + 4 sin (x)

Use the Pythagorean identity:

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

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

= - 5 + 4 (1 - sin^{2} (x)) + 4 sin (x)

Simplify

- 5 + 4 (1 - sin^{2} (x)) + 4 sin (x) : 4 sin (x) - 4 sin^{2} (x) - 1

- 5 + 4 (1 - sin^{2} (x)) + 4 sin (x)

Expand

4 (1 - sin^{2} (x)) : 4 - 4 sin^{2} (x)

4 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = 4, b = 1, c = sin^{2} (x)

= 4 \cdot 1 - 4 sin^{2} (x)

Multiply the numbers:

4 \cdot 1 = 4

= 4 - 4 sin^{2} (x)

= - 5 + 4 - 4 sin^{2} (x) + 4 sin (x)

Add/Subtract the numbers:

- 5 + 4 = - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

= 4 sin (x) - 4 sin^{2} (x) - 1

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

Solve by substitution

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

Let:

sin (x) = u

- 1 + 4 u - 4 u^{2} = 0

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

- 1 + 4 u - 4 u^{2} = 0

Write in the standard form

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

- 4 u^{2} + 4 u - 1 = 0

Solve with the quadratic formula

- 4 u^{2} + 4 u - 1 = 0

Quadratic Equation Formula:

For

a = - 4, b = 4, c = - 1

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

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

4^{2} - 4 (- 4) (- 1) = 0

4^{2} - 4 (- 4) (- 1)

Apply rule

- (- a) = a

= 4^{2} - 4 \cdot 4 \cdot 1

Multiply the numbers:

4 \cdot 4 \cdot 1 = 16

= 4^{2} - 16

4^{2} = 16

= 16 - 16

Subtract the numbers:

16 - 16 = 0

= 0

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

u = \frac{- 4}{2 ( - 4 )}

\frac{- 4}{2 ( - 4 )} = \frac{1}{2}

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

Remove parentheses:

(- a) = - a

= \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{1}{2}

The solution to the quadratic equation is:

u = \frac{1}{2}

Substitute back

u = sin (x)

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2}

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = \frac{1}{2}

General solutions for

sin (x) = \frac{1}{2}

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)+cos^2(x)= 5/4 ?
The general solution for sin(x)+cos^2(x)= 5/4 is x= pi/6+2pin,x=(5pi)/6+2pin