Sin(3x + 10) = cos (x)

In general cos x = cos a has solutions x = ±a + 360°k, integer k

sin(3x + 10) = cos x

Assuming that's 10°

cos(90 - (3x+10)) = cos x

cos(80 - 3x) = cos x

80 - 3x = ±x + 360°k

Plus first,

80-3x = x + 360°k

-4x = -80 +  360°k

x =20° - 90°k

Minus next,

80 - 3x = -x + 360°k

-2x = -80 + 360°k

x = 40° - 180°k

k runs over all the integers so we can flip the sign.

x = 20° + 90°k or x = 40° + 180°k, integer k

c. (3,2,6) is the answer.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°