What is the general solution for cos(x)= 1/(sqrt(37)) ?

The general solution for cos(x)= 1/(sqrt(37)) is x=1.40564…+2pin,x=2pi-1.40564…+2pin

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cos(x)= 1/(sqrt(37))

cos (x) = \frac{1}{37}

x = 1.40564 \dots + 2 πn, x = 2 π - 1.40564 \dots + 2 πn

Radians

x = 1.40564 \dots + 6.28318 \dots n, x = 4.87753 \dots + 6.28318 \dots n

Solution steps

cos (x) = \frac{1}{37}

Simplify

\frac{1}{37} : \frac{37}{37}

\frac{1}{37}

Multiply by the conjugate

\frac{37}{37}

= \frac{1 \cdot 37}{37 37}

1 \cdot 37 = 37

3737 = 37

3737

Apply radical rule:

a a = a

3737 = 37

= 37

= \frac{37}{37}

cos (x) = \frac{37}{37}

Apply trig inverse properties

cos (x) = \frac{37}{37}

General solutions for

cos (x) = \frac{37}{37}

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

x = arccos (\frac{37}{37}) + 2 πn, x = 2 π - arccos (\frac{37}{37}) + 2 πn

x = arccos (\frac{37}{37}) + 2 πn, x = 2 π - arccos (\frac{37}{37}) + 2 πn

Show solutions in decimal form

x = 1.40564 \dots + 2 πn, x = 2 π - 1.40564 \dots + 2 πn

\frac{sin ( 2 x )}{cos ( x )} = 4 + sin (2 x) \cdot tan (x)

sin (x) = 0.2174

cos (7 6^{\circ}) sec (8 x + 4) = 1

x, arccot (x^{2}) y = arccot (x^{2})

\frac{50}{33} = \frac{sin ( x )}{sin ( 12 0 ^{\circ} - x )}

What is the general solution for cos(x)= 1/(sqrt(37)) ?
The general solution for cos(x)= 1/(sqrt(37)) is x=1.40564…+2pin,x=2pi-1.40564…+2pin