What is the general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 ?

The general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 is x=1.23095…,x=360-1.23095…,x=120,x=240

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6cos^2(x)+cos(x)=1,0<= x<= 360

6 cos^{2} (x) + cos (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

x = 1.23095 \dots, x = 36 0^{\circ} - 1.23095 \dots, x = 12 0^{\circ}, x = 24 0^{\circ}

Solution steps

6 cos^{2} (x) + cos (x) = 1, 0^{\circ} \leq x \leq 36 0^{\circ}

Solve by substitution

cos (x) = \frac{1}{3}, cos (x) = - \frac{1}{2}

cos (x) = \frac{1}{3}, 0 \leq x \leq 36 0^{\circ} : x = arccos (\frac{1}{3}), x = 36 0^{\circ} - arccos (\frac{1}{3})

cos (x) = - \frac{1}{2}, 0 \leq x \leq 36 0^{\circ} : x = 12 0^{\circ}, x = 24 0^{\circ}

Combine all the solutions

x = arccos (\frac{1}{3}), x = 36 0^{\circ} - arccos (\frac{1}{3}), x = 12 0^{\circ}, x = 24 0^{\circ}

Show solutions in decimal form

x = 1.23095 \dots, x = 36 0^{\circ} - 1.23095 \dots, x = 12 0^{\circ}, x = 24 0^{\circ}

tan (y) = \frac{π}{4}

2 sin (x - 1) + 1 = 0

- 6 sin^{2} (x) + sin (x) = - 3

csc (x) + 2 = 0, 0 \leq x \leq 2 π

tan (x) = 3.7

What is the general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 ?
The general solution for 6cos^2(x)+cos(x)=1,0<= x<= 360 is x=1.23095…,x=360-1.23095…,x=120,x=240