What is the general solution for sqrt(7)*sin^2(x)+cos^2(x)-1=6sin(x) ?

The general solution for sqrt(7)*sin^2(x)+cos^2(x)-1=6sin(x) is x=2pin,x=pi+2pin

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sqrt(7)*sin^2(x)+cos^2(x)-1=6sin(x)

Solution

7 \cdot sin^{2} (x) + cos^{2} (x) - 1 = 6 sin (x)

Solution

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

7 sin^{2} (x) + cos^{2} (x) - 1 = 6 sin (x)

Subtract

6 sin (x)

from both sides

7 sin^{2} (x) + cos^{2} (x) - 1 - 6 sin (x) = 0

Rewrite using trig identities

- 1 + cos^{2} (x) - 6 sin (x) + sin^{2} (x) 7

Use the Pythagorean identity:

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

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

= - 6 sin (x) + sin^{2} (x) 7 - sin^{2} (x)

- sin^{2} (x) - 6 sin (x) + sin^{2} (x) 7 = 0

Solve by substitution

- sin^{2} (x) - 6 sin (x) + sin^{2} (x) 7 = 0

Let:

sin (x) = u

- u^{2} - 6 u + u^{2} 7 = 0

- u^{2} - 6 u + u^{2} 7 = 0 : u = 1 + 7, u = 0

- u^{2} - 6 u + u^{2} 7 = 0

Write in the standard form

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

(- 1 + 7) u^{2} - 6 u = 0

Solve with the quadratic formula

(- 1 + 7) u^{2} - 6 u = 0

Quadratic Equation Formula:

For

a = - 1 + 7, b = - 6, c = 0

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 + 7 ) \cdot 0}{2 ( - 1 + 7 )}

u_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 ( - 1 + 7 ) \cdot 0}{2 ( - 1 + 7 )}

(- 6)^{2} - 4 (- 1 + 7) \cdot 0 = 6

(- 6)^{2} - 4 (- 1 + 7) \cdot 0

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 6)^{2} = 6^{2}

= 6^{2} - 4 \cdot 0 \cdot (7 - 1)

Apply rule

0 \cdot a = 0

= 6^{2} - 0

6^{2} - 0 = 6^{2}

= 6^{2}

Apply radical rule:

assuming

a \geq 0

= 6

u_{1, 2} = \frac{- ( - 6 ) \pm 6}{2 ( - 1 + 7 )}

Separate the solutions

u_{1} = \frac{- ( - 6 ) + 6}{2 ( - 1 + 7 )}, u_{2} = \frac{- ( - 6 ) - 6}{2 ( - 1 + 7 )}

u = \frac{- ( - 6 ) + 6}{2 ( - 1 + 7 )} : 1 + 7

\frac{- ( - 6 ) + 6}{2 ( - 1 + 7 )}

Apply rule

- (- a) = a

= \frac{6 + 6}{2 ( - 1 + 7 )}

Add the numbers:

6 + 6 = 12

= \frac{12}{2 ( 7 - 1 )}

Divide the numbers:

\frac{12}{2} = 6

= \frac{6}{( - 1 + 7 )}

Remove parentheses:

(- a) = - a

= \frac{6}{- 1 + 7}

Rationalize

\frac{6}{- 1 + 7} : 1 + 7

\frac{6}{- 1 + 7}

Multiply by the conjugate

\frac{1 + 7}{1 + 7}

= \frac{6 ( 1 + 7 )}{( - 1 + 7 ) ( 1 + 7 )}

(- 1 + 7) (1 + 7) = 6

(- 1 + 7) (1 + 7)

Apply Difference of Two Squares Formula:

(a - b) (a + b) = a^{2} - b^{2}

a = 7, b = 1

= (7)^{2} - 1^{2}

Simplify

(7)^{2} - 1^{2} : 6

(7)^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (7)^{2} - 1

(7)^{2} = 7

(7)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (7^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 7^{\frac{1}{2} \cdot 2}

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 7

= 7 - 1

Subtract the numbers:

7 - 1 = 6

= 6

= 6

= \frac{6 ( 1 + 7 )}{6}

Divide the numbers:

\frac{6}{6} = 1

= 1 + 7

= 1 + 7

u = \frac{- ( - 6 ) - 6}{2 ( - 1 + 7 )} : 0

\frac{- ( - 6 ) - 6}{2 ( - 1 + 7 )}

Apply rule

- (- a) = a

= \frac{6 - 6}{2 ( - 1 + 7 )}

Subtract the numbers:

6 - 6 = 0

= \frac{0}{2 ( 7 - 1 )}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = 1 + 7, u = 0

Substitute back

u = sin (x)

sin (x) = 1 + 7, sin (x) = 0

sin (x) = 1 + 7, sin (x) = 0

sin (x) = 1 + 7 :

No Solution

sin (x) = 1 + 7

- 1 \leq sin (x) \leq 1

No Solution

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn

Popular Examples

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4 sec^{2} (x) - 3 tan^{2} (x) = 5 tan^{2} (x)

cos^3(x)=66

cos^{3} (x) = 66

2sqrt(3)*sin(4x+60^0)-3=0

23 \cdot sin (4 x + 6 0^{0}) - 3 = 0

(sin(x)+sin^2(x))/2 =0.5

\frac{sin ( x ) + sin ^{2} ( x )}{2} = 0.5

cos(b)= 3/5

cos (b) = \frac{3}{5}

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(7)*sin^2(x)+cos^2(x)-1=6sin(x) ?
The general solution for sqrt(7)*sin^2(x)+cos^2(x)-1=6sin(x) is x=2pin,x=pi+2pin