What is the general solution for 3sin(2x)=5cos^2(2x) ?

The general solution for 3sin(2x)=5cos^2(2x) is x=(0.83908…)/2+pin,x= pi/2-(0.83908…)/2+pin

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3sin(2x)=5cos^2(2x)

3 sin (2 x) = 5 cos^{2} (2 x)

x = \frac{0.83908 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.83908 \dots}{2} + πn

Solution steps

3 sin (2 x) = 5 cos^{2} (2 x)

Subtract

5 cos^{2} (2 x)

from both sides

3 sin (2 x) - 5 cos^{2} (2 x) = 0

Rewrite using trig identities

- (1 - sin^{2} (2 x)) \cdot 5 + 3 sin (2 x) = 0

Solve by substitution

sin (2 x) = \frac{- 3 + 109}{10}, sin (2 x) = \frac{- 3 - 109}{10}

sin (2 x) = \frac{- 3 + 109}{10} : x = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

sin (2 x) = \frac{- 3 - 109}{10} :

No Solution

Combine all the solutions

x = \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{- 3 + 109}{10} )}{2} + πn

Show solutions in decimal form

x = \frac{0.83908 \dots}{2} + πn, x = \frac{π}{2} - \frac{0.83908 \dots}{2} + πn

x, y^{n} + 25 y = cos (5 x)

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

sin^{2} (x) + sin (x) = 3 sin^{2} (x)

5 - 6 cos (θ) = 0

cos (x + \frac{π}{6}) cos (x - \frac{π}{6}) = cos (2 x)

What is the general solution for 3sin(2x)=5cos^2(2x) ?
The general solution for 3sin(2x)=5cos^2(2x) is x=(0.83908…)/2+pin,x= pi/2-(0.83908…)/2+pin