What is the general solution for cos(x)=((7^2+10^2-6^2)/(2(10)(6))) ?

The general solution for cos(x)=((7^2+10^2-6^2)/(2(10)(6))) is x=0.34324…+2pin,x=2pi-0.34324…+2pin

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cos(x)=((7^2+10^2-6^2)/(2(10)(6)))

cos (x) = (\frac{7 ^{2} + 1 0 ^{2} - 6 ^{2}}{2 ( 10 ) ( 6 )})

x = 0.34324 \dots + 2 πn, x = 2 π - 0.34324 \dots + 2 πn

Radians

x = 0.34324 \dots + 6.28318 \dots n, x = 5.93993 \dots + 6.28318 \dots n

Solution steps

cos (x) = (\frac{7 ^{2} + 1 0 ^{2} - 6 ^{2}}{2 ( 10 ) ( 6 )})

Simplify

\frac{7 ^{2} + 1 0 ^{2} - 6 ^{2}}{2 \cdot 10 \cdot 6} : \frac{113}{120}

\frac{7 ^{2} + 1 0 ^{2} - 6 ^{2}}{2 \cdot 10 \cdot 6}

7^{2} + 1 0^{2} - 6^{2} = 113

7^{2} + 1 0^{2} - 6^{2}

7^{2} = 49

= 49 + 1 0^{2} - 6^{2}

1 0^{2} = 100

= 49 + 100 - 6^{2}

6^{2} = 36

= 49 + 100 - 36

Add/Subtract the numbers:

49 + 100 - 36 = 113

= 113

= \frac{113}{2 \cdot 10 \cdot 6}

Multiply the numbers:

2 \cdot 10 \cdot 6 = 120

= \frac{113}{120}

Apply trig inverse properties

cos (x) = \frac{113}{120}

General solutions for

cos (x) = \frac{113}{120}

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

x = arccos (\frac{113}{120}) + 2 πn, x = 2 π - arccos (\frac{113}{120}) + 2 πn

x = arccos (\frac{113}{120}) + 2 πn, x = 2 π - arccos (\frac{113}{120}) + 2 πn

Show solutions in decimal form

x = 0.34324 \dots + 2 πn, x = 2 π - 0.34324 \dots + 2 πn

cos (2 x) = 3

tan (θ) = (\frac{2.5}{1.5})^{\frac{1}{3}}

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

sin (x) = 0.809

\frac{sin ( x )}{700} = \frac{sin ( 4 5 ^{\circ} )}{497}

What is the general solution for cos(x)=((7^2+10^2-6^2)/(2(10)(6))) ?
The general solution for cos(x)=((7^2+10^2-6^2)/(2(10)(6))) is x=0.34324…+2pin,x=2pi-0.34324…+2pin