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Popular >

(1+sin^2(x))=(4cos(x))^2

Solution

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

Solution

x = 1.22069 \dots + 2 πn, x = π - 1.22069 \dots + 2 πn, x = - 1.22069 \dots + 2 πn, x = π + 1.22069 \dots + 2 πn

Degrees

x = 69.94041 \dots^{\circ} + 36 0^{\circ} n, x = 110.05958 \dots^{\circ} + 36 0^{\circ} n, x = - 69.94041 \dots^{\circ} + 36 0^{\circ} n, x = 249.94041 \dots^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

(4 cos (x))^{2}

from both sides

1 + sin^{2} (x) - 16 cos^{2} (x) = 0

Rewrite using trig identities

1 + sin^{2} (x) - 16 cos^{2} (x)

Use the Pythagorean identity:

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

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

= 1 + sin^{2} (x) - 16 (1 - sin^{2} (x))

Simplify

1 + sin^{2} (x) - 16 (1 - sin^{2} (x)) : 17 sin^{2} (x) - 15

1 + sin^{2} (x) - 16 (1 - sin^{2} (x))

Expand

- 16 (1 - sin^{2} (x)) : - 16 + 16 sin^{2} (x)

- 16 (1 - sin^{2} (x))

Apply the distributive law:

a (b - c) = ab - a c

a = - 16, b = 1, c = sin^{2} (x)

= - 16 \cdot 1 - (- 16) sin^{2} (x)

Apply minus-plus rules

- (- a) = a

= - 16 \cdot 1 + 16 sin^{2} (x)

Multiply the numbers:

16 \cdot 1 = 16

= - 16 + 16 sin^{2} (x)

= 1 + sin^{2} (x) - 16 + 16 sin^{2} (x)

Simplify

1 + sin^{2} (x) - 16 + 16 sin^{2} (x) : 17 sin^{2} (x) - 15

1 + sin^{2} (x) - 16 + 16 sin^{2} (x)

Group like terms

= sin^{2} (x) + 16 sin^{2} (x) + 1 - 16

Add similar elements:

sin^{2} (x) + 16 sin^{2} (x) = 17 sin^{2} (x)

= 17 sin^{2} (x) + 1 - 16

Add/Subtract the numbers:

1 - 16 = - 15

= 17 sin^{2} (x) - 15

= 17 sin^{2} (x) - 15

= 17 sin^{2} (x) - 15

- 15 + 17 sin^{2} (x) = 0

Solve by substitution

- 15 + 17 sin^{2} (x) = 0

Let:

sin (x) = u

- 15 + 17 u^{2} = 0

- 15 + 17 u^{2} = 0 : u = \frac{15}{17}, u = - \frac{15}{17}

- 15 + 17 u^{2} = 0

Move

15

to the right side

- 15 + 17 u^{2} = 0

Add

15

to both sides

- 15 + 17 u^{2} + 15 = 0 + 15

Simplify

17 u^{2} = 15

17 u^{2} = 15

Divide both sides by

17

17 u^{2} = 15

Divide both sides by

17

\frac{17 u ^{2}}{17} = \frac{15}{17}

Simplify

u^{2} = \frac{15}{17}

u^{2} = \frac{15}{17}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{15}{17}, u = - \frac{15}{17}

Substitute back

u = sin (x)

sin (x) = \frac{15}{17}, sin (x) = - \frac{15}{17}

sin (x) = \frac{15}{17}, sin (x) = - \frac{15}{17}

sin (x) = \frac{15}{17} : x = arcsin (\frac{15}{17}) + 2 πn, x = π - arcsin (\frac{15}{17}) + 2 πn

sin (x) = \frac{15}{17}

Apply trig inverse properties

sin (x) = \frac{15}{17}

General solutions for

sin (x) = \frac{15}{17}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

x = arcsin (\frac{15}{17}) + 2 πn, x = π - arcsin (\frac{15}{17}) + 2 πn

x = arcsin (\frac{15}{17}) + 2 πn, x = π - arcsin (\frac{15}{17}) + 2 πn

sin (x) = - \frac{15}{17} : x = arcsin (- \frac{15}{17}) + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

sin (x) = - \frac{15}{17}

Apply trig inverse properties

sin (x) = - \frac{15}{17}

General solutions for

sin (x) = - \frac{15}{17}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (- \frac{15}{17}) + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

x = arcsin (- \frac{15}{17}) + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

Combine all the solutions

x = arcsin (\frac{15}{17}) + 2 πn, x = π - arcsin (\frac{15}{17}) + 2 πn, x = arcsin (- \frac{15}{17}) + 2 πn, x = π + arcsin (\frac{15}{17}) + 2 πn

Show solutions in decimal form

x = 1.22069 \dots + 2 πn, x = π - 1.22069 \dots + 2 πn, x = - 1.22069 \dots + 2 πn, x = π + 1.22069 \dots + 2 πn

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