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Popular >

(cos(t)-4)(2sin^2(t)-1)=0

Solution

(cos (t) - 4) (2 sin^{2} (t) - 1) = 0

Solution

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

+1

Degrees

t = 4 5^{\circ} + 36 0^{\circ} n, t = 13 5^{\circ} + 36 0^{\circ} n, t = 22 5^{\circ} + 36 0^{\circ} n, t = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

(cos (t) - 4) (2 sin^{2} (t) - 1) = 0

Solving each part separately

cos (t) - 4 = 0 or 2 sin^{2} (t) - 1 = 0

cos (t) - 4 = 0 :

No Solution

cos (t) - 4 = 0

Move

4

to the right side

cos (t) - 4 = 0

Add

4

to both sides

cos (t) - 4 + 4 = 0 + 4

Simplify

cos (t) = 4

cos (t) = 4

- 1 \leq cos (x) \leq 1

No Solution

2 sin^{2} (t) - 1 = 0 : t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

2 sin^{2} (t) - 1 = 0

Solve by substitution

2 sin^{2} (t) - 1 = 0

Let:

sin (t) = u

2 u^{2} - 1 = 0

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

2 u^{2} - 1 = 0

Move

1

to the right side

2 u^{2} - 1 = 0

Add

1

to both sides

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

Simplify

2 u^{2} = 1

2 u^{2} = 1

Divide both sides by

2

2 u^{2} = 1

Divide both sides by

2

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

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

Substitute back

u = sin (t)

sin (t) = \frac{1}{2}, sin (t) = - \frac{1}{2}

sin (t) = \frac{1}{2}, sin (t) = - \frac{1}{2}

sin (t) = \frac{1}{2} : t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn

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

General solutions for

sin (t) = \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}

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn

sin (t) = - \frac{1}{2} : t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

sin (t) = - \frac{1}{2}

General solutions for

sin (t) = - \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}

t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

Combine all the solutions

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

Combine all the solutions

t = \frac{π}{4} + 2 πn, t = \frac{3 π}{4} + 2 πn, t = \frac{5 π}{4} + 2 πn, t = \frac{7 π}{4} + 2 πn

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