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

The general solution for 2cos^2(x)+sin(x)=1,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6 ,x= pi/2

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2cos^2(x)+sin(x)=1,0<= x<= 2pi

2 cos^{2} (x) + sin (x) = 1, 0 \leq x \leq 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{2}

Solution steps

2 cos^{2} (x) + sin (x) = 1, 0 \leq x \leq 2 π

Subtract

1

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{7 π}{6}, x = \frac{11 π}{6}

sin (x) = 1, 0 \leq x \leq 2 π : x = \frac{π}{2}

Combine all the solutions

x = \frac{7 π}{6}, x = \frac{11 π}{6}, x = \frac{π}{2}

cos (x) = \frac{5}{5}

4 sin (x) cos (x) = 2 sin (x)

(sec (t) - 2) (cot (t) + 1) = 0

tan (θ) = \frac{50}{60}

x, tan^{2} (x) = \frac{3}{2} sec (x)

What is the general solution for 2cos^2(x)+sin(x)=1,0<= x<= 2pi ?
The general solution for 2cos^2(x)+sin(x)=1,0<= x<= 2pi is x=(7pi)/6 ,x=(11pi)/6 ,x= pi/2