What is the general solution for 3cos(x)+sec(145)=1 ?

The general solution for 3cos(x)+sec(145)=1 is x=0.7373420134+360n,x=360-0.7373420134+360n

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3cos(x)+sec(145)=1

3 cos (x) + sec (14 5^{\circ}) = 1

x = 0.7373420134 + 36 0^{\circ} n, x = 36 0^{\circ} - 0.7373420134 + 36 0^{\circ} n

Radians

x = 0.7373420134 + 2 πn, x = 2 π - 0.7373420134 + 2 πn

Solution steps

3 cos (x) + sec (14 5^{\circ}) = 1

Move

sec (14 5^{\circ})

to the right side

3 cos (x) + sec (14 5^{\circ}) = 1

Subtract

sec (14 5^{\circ})

from both sides

3 cos (x) + sec (14 5^{\circ}) - sec (14 5^{\circ}) = 1 - sec (14 5^{\circ})

Simplify

3 cos (x) = 1 - sec (14 5^{\circ})

3 cos (x) = 1 - sec (14 5^{\circ})

Divide both sides by

3

3 cos (x) = 1 - sec (14 5^{\circ})

Divide both sides by

3

\frac{3 cos ( x )}{3} = \frac{1}{3} - \frac{sec ( 14 5 ^{\circ} )}{3}

Simplify

\frac{3 cos ( x )}{3} = \frac{1}{3} - \frac{sec ( 14 5 ^{\circ} )}{3}

Simplify

\frac{3 cos ( x )}{3} : cos (x)

\frac{3 cos ( x )}{3}

Divide the numbers:

\frac{3}{3} = 1

= cos (x)

Simplify

\frac{1}{3} - \frac{sec ( 14 5 ^{\circ} )}{3} : \frac{1 - sec ( 14 5 ^{\circ} )}{3}

\frac{1}{3} - \frac{sec ( 14 5 ^{\circ} )}{3}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{1 - sec ( 14 5 ^{\circ} )}{3}

cos (x) = \frac{1 - sec ( 14 5 ^{\circ} )}{3}

cos (x) = \frac{1 - sec ( 14 5 ^{\circ} )}{3}

cos (x) = \frac{1 - sec ( 14 5 ^{\circ} )}{3}

Apply trig inverse properties

cos (x) = \frac{1 - sec ( 14 5 ^{\circ} )}{3}

General solutions for

cos (x) = \frac{1 - sec ( 14 5 ^{\circ} )}{3}

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

x = arccos (\frac{1 - sec ( 14 5 ^{\circ} )}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1 - sec ( 14 5 ^{\circ} )}{3}) + 36 0^{\circ} n

x = arccos (\frac{1 - sec ( 14 5 ^{\circ} )}{3}) + 36 0^{\circ} n, x = 36 0^{\circ} - arccos (\frac{1 - sec ( 14 5 ^{\circ} )}{3}) + 36 0^{\circ} n

Show solutions in decimal form

x = 0.7373420134 + 36 0^{\circ} n, x = 36 0^{\circ} - 0.7373420134 + 36 0^{\circ} n

sec^{2} (x) + 3 tan^{2} (x) = 13

1609.3 = \frac{1.13 3 ^{2}}{( 9.8 ) tan ( θ )}

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

cos (x) = (\frac{- 3}{4})

4 (cos (x))^{2} = 5 - 4 sin (x)

What is the general solution for 3cos(x)+sec(145)=1 ?
The general solution for 3cos(x)+sec(145)=1 is x=0.7373420134+360n,x=360-0.7373420134+360n