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Popular >

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

Solution

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

Solution

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

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Solution steps

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

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

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

Combine all the solutions

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

Popular Examples

cot(x)=-1,0<= x<= 2pi

cot (x) = - 1, 0 \leq x \leq 2 π

(2cos(x)-sqrt(2))(2sin(x)+sqrt(2))=0

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

4 sec^{2} (x) - 3 = 0

cos^2(x)= 1/4 ,0<= x<= 2pi

cos^{2} (x) = \frac{1}{4}, 0 \leq x \leq 2 π

sin (θ) = - \frac{4}{9}

Frequently Asked Questions (FAQ)

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