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Popular >

2sin^2(2θ)=1+sin(2θ)

Solution

2 sin^{2} (2 θ) = 1 + sin (2 θ)

Solution

θ = \frac{π}{4} + πn, θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

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Solution steps

2 sin^{2} (2 θ) = 1 + sin (2 θ)

Solve by substitution

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

sin (2 θ) = 1 : θ = \frac{π}{4} + πn

sin (2 θ) = - \frac{1}{2} : θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Combine all the solutions

θ = \frac{π}{4} + πn, θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

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sin (2 x) - cos (x) = 0, 0 \leq x \leq 2 π

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csc (θ) = - \frac{2}{3}

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

25sin(x)cos(x)-5sin(x)+20cos(x)=4

25 sin (x) cos (x) - 5 sin (x) + 20 cos (x) = 4

Frequently Asked Questions (FAQ)

What is the general solution for 2sin^2(2θ)=1+sin(2θ) ?
The general solution for 2sin^2(2θ)=1+sin(2θ) is θ= pi/4+pin,θ=(7pi)/(12)+pin,θ=(11pi)/(12)+pin