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Popular >

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

Solution

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

Solution

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

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Solution steps

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

Solve by substitution

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

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

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

Combine all the solutions

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

Popular Examples

15tan(θ)-7=5tan(θ)-3

15 tan (θ) - 7 = 5 tan (θ) - 3

sin^2(x)-9cos(x)+9=0,0<= x<= 2pi

sin^{2} (x) - 9 cos (x) + 9 = 0, 0 \leq x \leq 2 π

-10cos(2x)-32cos(x)-22=0

- 10 cos (2 x) - 32 cos (x) - 22 = 0

solvefor θ,sin(θ)=-3/4 solve for

θ, sin (θ) = - \frac{3}{4}

tan(x)*sec(x)=sqrt(2)

tan (x) \cdot sec (x) = 2

Frequently Asked Questions (FAQ)

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