What is the general solution for 7cos(b)-sqrt(13)=0 ?

The general solution for 7cos(b)-sqrt(13)=0 is b=1.02969…+2pin,b=2pi-1.02969…+2pin

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7cos(b)-sqrt(13)=0

7 cos (b) - 13 = 0

b = 1.02969 \dots + 2 πn, b = 2 π - 1.02969 \dots + 2 πn

Degrees

b = 58.99728 \dots^{\circ} + 36 0^{\circ} n, b = 301.00271 \dots^{\circ} + 36 0^{\circ} n

Solution steps

7 cos (b) - 13 = 0

Move

13

to the right side

7 cos (b) - 13 = 0

Add

13

to both sides

7 cos (b) - 13 + 13 = 0 + 13

Simplify

7 cos (b) = 13

7 cos (b) = 13

Divide both sides by

7

7 cos (b) = 13

Divide both sides by

7

\frac{7 cos ( b )}{7} = \frac{13}{7}

Simplify

cos (b) = \frac{13}{7}

cos (b) = \frac{13}{7}

Apply trig inverse properties

cos (b) = \frac{13}{7}

General solutions for

cos (b) = \frac{13}{7}

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

b = arccos (\frac{13}{7}) + 2 πn, b = 2 π - arccos (\frac{13}{7}) + 2 πn

b = arccos (\frac{13}{7}) + 2 πn, b = 2 π - arccos (\frac{13}{7}) + 2 πn

Show solutions in decimal form

b = 1.02969 \dots + 2 πn, b = 2 π - 1.02969 \dots + 2 πn

cos (θ) = \frac{6}{3.92739}

sin (x) = - \frac{1}{3}, 0 \leq x \leq 2 π

3 csc (x) = 2 tan (x)

3 = 6 cos (x)

2 cos (a) \cdot sin (2 a) - 1 + 2 sin (2 a) - cos (a) = 0

What is the general solution for 7cos(b)-sqrt(13)=0 ?
The general solution for 7cos(b)-sqrt(13)=0 is b=1.02969…+2pin,b=2pi-1.02969…+2pin