What is the general solution for sin(10)=2sin(a)cos(a) ?

The general solution for sin(10)=2sin(a)cos(a) is a=5+180n,a=90-5+180n

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sin(10)=2sin(a)cos(a)

sin (1 0^{\circ}) = 2 sin (a) cos (a)

a = 5^{\circ} + 18 0^{\circ} n, a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

Solution steps

sin (1 0^{\circ}) = 2 sin (a) cos (a)

Subtract

2 sin (a) cos (a)

from both sides

sin (1 0^{\circ}) - 2 sin (a) cos (a) = 0

Rewrite using trig identities

sin (1 0^{\circ}) - sin (2 a) = 0

Move

sin (1 0^{\circ})

to the right side

- sin (2 a) = - sin (1 0^{\circ})

Divide both sides by

- 1

sin (2 a) = sin (1 0^{\circ})

Apply trig inverse properties

2 a = arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n, 2 a = 18 0^{\circ} - arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n

Solve

2 a = arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n : a = 5^{\circ} + 18 0^{\circ} n

Solve

2 a = 18 0^{\circ} - arcsin (sin (1 0^{\circ})) + 36 0^{\circ} n : a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

a = 5^{\circ} + 18 0^{\circ} n, a = 9 0^{\circ} - 5^{\circ} + 18 0^{\circ} n

cos^{2} (x) = 2 + 2 sin (x), 0 \leq x \leq 2 π

cos^{2} (x) = \frac{3}{5}

1 - sec^{2} (x) = tan^{2} (x)

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

- 2 cos^{2} (x) + 3 cos (x) = 1

What is the general solution for sin(10)=2sin(a)cos(a) ?
The general solution for sin(10)=2sin(a)cos(a) is a=5+180n,a=90-5+180n