What is the general solution for cos(b)=(cos(80))/(cos(50)) ?

The general solution for cos(b)=(cos(80))/(cos(50)) is b=1.29724…+360n,b=360-1.29724…+360n

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cos(b)=(cos(80))/(cos(50))

cos (b) = \frac{cos ( 8 0 ^{\circ} )}{cos ( 5 0 ^{\circ} )}

b = 1.29724 \dots + 36 0^{\circ} n, b = 36 0^{\circ} - 1.29724 \dots + 36 0^{\circ} n

Radians

b = 1.29724 \dots + 2 πn, b = 2 π - 1.29724 \dots + 2 πn

Solution steps

cos (b) = \frac{cos ( 8 0 ^{\circ} )}{cos ( 5 0 ^{\circ} )}

Apply trig inverse properties

cos (b) = \frac{cos ( 8 0 ^{\circ} )}{cos ( 5 0 ^{\circ} )}

General solutions for

cos (b) = \frac{cos ( 8 0 ^{\circ} )}{cos ( 5 0 ^{\circ} )}

cos (x) = a \Rightarrow x = arccos (a) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (a) + 36 0^{\circ} n

b = arccos (\frac{cos ( 8 0 ^{\circ} )}{cos ( 5 0 ^{\circ} )}) + 36 0^{\circ} n, b = 36 0^{\circ} - arccos (\frac{cos ( 8 0 ^{\circ} )}{cos ( 5 0 ^{\circ} )}) + 36 0^{\circ} n

b = arccos (\frac{cos ( 8 0 ^{\circ} )}{cos ( 5 0 ^{\circ} )}) + 36 0^{\circ} n, b = 36 0^{\circ} - arccos (\frac{cos ( 8 0 ^{\circ} )}{cos ( 5 0 ^{\circ} )}) + 36 0^{\circ} n

Show solutions in decimal form

b = 1.29724 \dots + 36 0^{\circ} n, b = 36 0^{\circ} - 1.29724 \dots + 36 0^{\circ} n

2 sec^{2} (x) tan (x) - 2 sec^{2} (x) = 0

sin (x) = \frac{10}{18}

tan (2 x) = 2 cos (x)

3 sec^{2} (θ) - 2 tan^{2} (θ) - 4 = 0

cos^{2} (x) + 5 sin^{2} (x) = 2

What is the general solution for cos(b)=(cos(80))/(cos(50)) ?
The general solution for cos(b)=(cos(80))/(cos(50)) is b=1.29724…+360n,b=360-1.29724…+360n