( 2\sin x = 1 ) ( \cos^2 x - \sin^2 x = 0 ) ( \tan(2x) = \sqrt3 )
Let ( t = 2x ). Solve ( \tan t = 1 ). Principal value: ( t = \pi/4 ). Tangent period is ( \pi ): ( t = \pi/4 + k\pi ). Thus ( 2x = \pi/4 + k\pi \Rightarrow x = \pi/8 + k\pi/2 ). ecuaciones trigonometricas 1 bachillerato
Method 1: Divide by ( \cos x ) (if ( \cos x \neq 0 )): ( \tan x = 1 \Rightarrow x = \pi/4 + k\pi ). Check ( \cos x = 0 ) gives ( x = \pi/2, 3\pi/2 ), but those don’t satisfy original (sine ≠ cosine). So fine. ( 2\sin x = 1 ) ( \cos^2
With practice, solving trigonometric equations becomes systematic. Memorize the general solution forms and always check your solutions in the original equation. Tangent period is ( \pi ): ( t = \pi/4 + k\pi )
Bring all terms: ( \sin x \cos x - \frac12\sin x = 0 ) ( \sin x (\cos x - 1/2) = 0 )