What is the general solution for 2cos(x)+1=sin^2(x) ?

The general solution for 2cos(x)+1=sin^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n, x = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

sin^{2} (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Solve by substitution

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

Let:

cos (x) = u

u^{2} + 2 u = 0

u^{2} + 2 u = 0 : u = 0, u = - 2

u^{2} + 2 u = 0

Solve with the quadratic formula

u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = 1, b = 2, c = 0

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 0 = 2

2^{2} - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 2 + 2 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 2 - 2}{2 \cdot 1} : - 2

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

Subtract the numbers:

- 2 - 2 = - 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = 0, u = - 2

Substitute back

u = cos (x)

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

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

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = 0

General solutions for

cos (x) = 0

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}

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

cos (x) = - 2 :

No Solution

cos (x) = - 2

- 1 \leq cos (x) \leq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

Popular Examples

4cos^2(x)-11cos(x)+7=0

4 cos^{2} (x) - 11 cos (x) + 7 = 0

sqrt(3)tan(x)+2sec(x)=1

3 tan (x) + 2 sec (x) = 1

cos(a)=0.14

cos (a) = 0.14

(1+cos^2(x))/(sin^2(x))= 1/(sin^2(x))

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

7sin(a)= 1/11

7 sin (a) = \frac{1}{11}

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(x)+1=sin^2(x) ?
The general solution for 2cos(x)+1=sin^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin