What is the general solution for cos(5a+40)= 3/5 ?

The general solution for cos(5a+40)= 3/5 is a=(360n)/5-8+(0.92729…)/5 ,a=72-8+(360n)/5-(0.92729…)/5

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cos(5a+40)= 3/5

cos (5 a + 4 0^{\circ}) = \frac{3}{5}

a = \frac{36 0 ^{\circ} n}{5} - 8^{\circ} + \frac{0.92729 \dots}{5}, a = 7 2^{\circ} - 8^{\circ} + \frac{36 0 ^{\circ} n}{5} - \frac{0.92729 \dots}{5}

Solution steps

cos (5 a + 4 0^{\circ}) = \frac{3}{5}

Apply trig inverse properties

5 a + 4 0^{\circ} = arccos (\frac{3}{5}) + 36 0^{\circ} n, 5 a + 4 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n

Solve

5 a + 4 0^{\circ} = arccos (\frac{3}{5}) + 36 0^{\circ} n : a = \frac{36 0 ^{\circ} n}{5} - 8^{\circ} + \frac{arccos ( \frac{3}{5} )}{5}

Solve

5 a + 4 0^{\circ} = 36 0^{\circ} - arccos (\frac{3}{5}) + 36 0^{\circ} n : a = 7 2^{\circ} - 8^{\circ} + \frac{36 0 ^{\circ} n}{5} - \frac{arccos ( \frac{3}{5} )}{5}

a = \frac{36 0 ^{\circ} n}{5} - 8^{\circ} + \frac{arccos ( \frac{3}{5} )}{5}, a = 7 2^{\circ} - 8^{\circ} + \frac{36 0 ^{\circ} n}{5} - \frac{arccos ( \frac{3}{5} )}{5}

Show solutions in decimal form

a = \frac{36 0 ^{\circ} n}{5} - 8^{\circ} + \frac{0.92729 \dots}{5}, a = 7 2^{\circ} - 8^{\circ} + \frac{36 0 ^{\circ} n}{5} - \frac{0.92729 \dots}{5}

2 cos (x^{2}) - 2 = 0

x, sin (x) = - 2 \cdot cos (x)

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

2 cos (t) - 1 = 0, 0 <, = t \leq 2 π

2 sin (2 (θ + \frac{π}{6})) = - 2

What is the general solution for cos(5a+40)= 3/5 ?
The general solution for cos(5a+40)= 3/5 is a=(360n)/5-8+(0.92729…)/5 ,a=72-8+(360n)/5-(0.92729…)/5