What is the general solution for 4cos(x)=sec(x)+3,0<= x<2pi ?

The general solution for 4cos(x)=sec(x)+3,0<= x<2pi is x=1.82347…,x=-1.82347…+2pi,x=0

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4cos(x)=sec(x)+3,0<= x<2pi

4 cos (x) = sec (x) + 3, 0 \leq x < 2 π

x = 1.82347 \dots, x = - 1.82347 \dots + 2 π, x = 0

Solution steps

4 cos (x) = sec (x) + 3, 0 \leq x < 2 π

Subtract

sec (x) + 3

from both sides

4 cos (x) - sec (x) - 3 = 0

Rewrite using trig identities

- 3 + \frac{4}{sec ( x )} - sec (x) = 0

Solve by substitution

sec (x) = - 4, sec (x) = 1

sec (x) = - 4, 0 \leq x < 2 π : x = arcsec (- 4), x = - arcsec (- 4) + 2 π

sec (x) = 1, 0 \leq x < 2 π : x = 0

Combine all the solutions

x = arcsec (- 4), x = - arcsec (- 4) + 2 π, x = 0

Show solutions in decimal form

x = 1.82347 \dots, x = - 1.82347 \dots + 2 π, x = 0

x, - \frac{1}{2 y ^{2}} = 3 sin (x) - \frac{1}{8}

sin (θ) = \frac{150 sin ( 11 5 ^{\circ} )}{212.6}

- sin (x) - sin (2 x) = 0

2 cos^{2} (2 x) + cos (2 x) - 1 = 0

cos (θ) = \frac{0}{8}

What is the general solution for 4cos(x)=sec(x)+3,0<= x<2pi ?
The general solution for 4cos(x)=sec(x)+3,0<= x<2pi is x=1.82347…,x=-1.82347…+2pi,x=0