What is the general solution for 3sin^2(2x)+7cos(2x)-3=0 ?

The general solution for 3sin^2(2x)+7cos(2x)-3=0 is x= pi/4+pin,x=(3pi)/4+pin

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3sin^2(2x)+7cos(2x)-3=0

3 sin^{2} (2 x) + 7 cos (2 x) - 3 = 0

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Solution steps

3 sin^{2} (2 x) + 7 cos (2 x) - 3 = 0

Rewrite using trig identities

- 3 cos^{2} (2 x) + 7 cos (2 x) = 0

Solve by substitution

cos (2 x) = 0, cos (2 x) = \frac{7}{3}

cos (2 x) = 0 : x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

cos (2 x) = \frac{7}{3} :

No Solution

Combine all the solutions

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

3 tan (2 θ) - 1 = 0

3 cos^{2} (θ) - 1 = 2 cos (θ)

cos (\frac{x}{2}) = sin (x)

4 sin^{2} (x) - 5 sin (x) + 1 = 0

tan (x) = - 0.6293

What is the general solution for 3sin^2(2x)+7cos(2x)-3=0 ?
The general solution for 3sin^2(2x)+7cos(2x)-3=0 is x= pi/4+pin,x=(3pi)/4+pin