What is the general solution for 0=1.5961*cos(x)+0.0422*cos(2*x) ?

The general solution for 0=1.5961*cos(x)+0.0422*cos(2*x) is x=1.54439…+2pin,x=2pi-1.54439…+2pin

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0=1.5961cos(x)+0.0422cos(2*x)

0 = 1.5961 \cdot cos (x) + 0.0422 \cdot cos (2 \cdot x)

x = 1.54439 \dots + 2 πn, x = 2 π - 1.54439 \dots + 2 πn

Solution steps

0 = 1.5961 cos (x) + 0.0422 cos (2 x)

Switch sides

1.5961 cos (x) + 0.0422 cos (2 x) = 0

Rewrite using trig identities

(- 1 + 2 cos^{2} (x)) \cdot 0.0422 + 1.5961 cos (x) = 0

Solve by substitution

cos (x) = \frac{- 1.5961 + 2.56178193}{0.1688}, cos (x) = \frac{- 1.5961 - 2.56178193}{0.1688}

cos (x) = \frac{- 1.5961 + 2.56178193}{0.1688} : x = arccos (\frac{- 1.5961 + 2.56178193}{0.1688}) + 2 πn, x = 2 π - arccos (\frac{- 1.5961 + 2.56178193}{0.1688}) + 2 πn

cos (x) = \frac{- 1.5961 - 2.56178193}{0.1688} :

No Solution

Combine all the solutions

x = arccos (\frac{- 1.5961 + 2.56178193}{0.1688}) + 2 πn, x = 2 π - arccos (\frac{- 1.5961 + 2.56178193}{0.1688}) + 2 πn

Show solutions in decimal form

x = 1.54439 \dots + 2 πn, x = 2 π - 1.54439 \dots + 2 πn

6 cos (7 θ) = 0

csc^{2} (θ) = 3 cot (θ) + 5

sin^{2} (x) - 2 sin (x) = 0, 0 \leq x < 2 π

cos (2 x) = \frac{( 5 k - 8 )}{5}

x, - f = sin (x)

What is the general solution for 0=1.5961*cos(x)+0.0422*cos(2*x) ?
The general solution for 0=1.5961*cos(x)+0.0422*cos(2*x) is x=1.54439…+2pin,x=2pi-1.54439…+2pin