Solve

Question

sin2x - 2cos2x=1 in radians? 

Answer 1

The trigonometric equation is sin (2x) - 2cos (2x) = 1.

Divide each side by cos (2x).

[sin 2x/cos 2x] - 2[cos 2x/cos 2x] = 1/cos 2x

tan (2x) - 2 = sec (2x)

Squaring on both sides.

[ tan (2x) - 2 ]^2 = sec^2 (2x)

tan^2 (2x) - 4tan (2x) + 4 = sec^2 (2x)

tan^2 (2x) - 4tan (2x) + 4 = 1 + tan^2 (2x)                                   ( ∴ sec^2 (2x) =  1 + tan^2 (2x) )

- 4tan (2x) + 4 = 1

4tan (2x) = 4 - 1 = 3

tan (2x) = 3/4

2x = tan^(- 1) (3/4)

x = 1/2(tan^(- 1) (3/4)).

The solution is 1/2[ πn + (tan^(- 1) (3/4)], where n € Z.

Answer 2

The tigonometric equation is sin(2x) - 2 cos(2x) = 1.

Apply double angle formulas : sin(2u) = 2 sin(u) cos(u) and cos(2u) = 1 - 2 sin²(u).

2 sin(x) cos(x) - 2[ 1 - 2 sin²(x) ] = 1

2 sin(x) cos(x) - 2 + 4 sin²(x) ] = 1

2 sin(x) cos(x) + 4 sin²(x) ] = 3

Divide each side by cos²(x).

[ 2 sin(x) cos(x) ] / [ cos²(x) ] + 4 sin²(x) ] / [ cos²(x) ] = 3 / [ cos²(x) ]

2 [ sin(x) / cos(x) ] + 4 [ sin²(x) / cos²(x) ] = 3 / [ cos²(x) ]

2 tan(x) + 4 tan²(x) = 3 sec²(x)

2 tan(x) + 4 tan²(x) = 3[ 1 + tan²(x) ]

2 tan(x) + 4 tan²(x) = 3 + 3 tan²(x)

tan²(x) + 2 tan(x) - 3 = 0

tan²(x) + 3 tan(x) - tan(x) - 3 = 0

tan(x) [ tan(x) + 3 ] - 1 [ tan(x) + 3 ] = 0

[ tan(x) - 1 ] [ tan(x) + 3 ] = 0

tan(x) = 1 and tan(x) = - 3

tan(x) = tan(π/4) and tan(x) = tan[ tan^{- 1}(- 3) ]

The genaral solution of tan(θ) = tan(α) is θ = nπ + α, where n is an integer.

x = nπ + π/4 and x = nπ + tan^{- 1}(- 3).

x = nπ + π/4 and x = nπ - tan^{- 1}(3).

The solutions are x = nπ + π/4 and x = nπ - tan^{- 1}(3), where n is an integer.