Ecuaciones Trigonometricas 1 Bachillerato Ejercicios Resueltos Fixed May 2026

Resolver: $2\cos^2(x) - 3\cos(x) + 1 = 0$

Solución:

  • Volvolvemos a la trigonometría:

    Solución General: $x = 2\pi n \quad ; \quad x = \pm \frac\pi3 + 2\pi n, \quad n \in \mathbbZ$.

    • We will present fixed, solved exercises for each major type.

    Solve: (\sin 2x - \sin x = 0).

    Step 1: Identity: (\sin 2x = 2\sin x \cos x).
    So (2\sin x \cos x - \sin x = 0 \Rightarrow \sin x (2\cos x - 1) = 0).

    Step 2:

    General: [ x = k\pi, \quad x = \frac\pi3 + 2k\pi, \quad x = \frac5\pi3 + 2k\pi. ] Resolver: $2\cos^2(x) - 3\cos(x) + 1 = 0$ Solución:

    Exercise 5: Solve ( 2\sin^2 x - \sin x - 1 = 0 ) for ( x \in [0, 2\pi) ).

    Step 1: Factor like a quadratic: Let ( y = \sin x ).
    ( 2y^2 - y - 1 = 0 \Rightarrow (2y + 1)(y - 1) = 0 )
    ( y = 1 ) or ( y = -\frac12 ).

    Step 2:
    If ( \sin x = 1 \Rightarrow x = \frac\pi2 ).
    If ( \sin x = -\frac12 ): reference ( \pi/6 ), sine negative in QIII and QIV.
    QIII: ( x = \pi + \frac\pi6 = \frac7\pi6 )
    QIV: ( x = 2\pi - \frac\pi6 = \frac11\pi6 ).

    Answer: ( x = \frac\pi2,\ \frac7\pi6,\ \frac11\pi6 ).

    Exercise 6: Solve ( \cos 2x = \cos x ) for ( x \in [0, 2\pi) ).

    Step 1: Use identity ( \cos 2x = 2\cos^2 x - 1 ).
    ( 2\cos^2 x - 1 = \cos x )
    ( 2\cos^2 x - \cos x - 1 = 0 )
    Let ( y = \cos x ): ( 2y^2 - y - 1 = 0 \Rightarrow (2y + 1)(y - 1) = 0 )
    ( y = 1 ) or ( y = -1/2 ).

    Step 2:
    ( \cos x = 1 \Rightarrow x = 0 )
    ( \cos x = -1/2 \Rightarrow x = \frac2\pi3,\ \frac4\pi3 ).
    Answer: ( 0,\ \frac2\pi3,\ \frac4\pi3 ).

    Enunciado: Resuelve ( \sqrt3 \tg x - 1 = 0 ) Volvolvemos a la trigonometría:

    Solución (Fixed):

    Resuelve la siguiente ecuación: $$2\cos^2 x - 3\cos x + 1 = 0$$

    Solución:

  • Volver a la variable original ($x$):

    Caso A: $t_1 = 1 \Rightarrow \cos x = 1$ El ángulo cuyo coseno es 1 es $0^\circ$. $$x = 0^\circ + 360^\circ \cdot k$$

    Caso B: $t_2 = \frac12 \Rightarrow \cos x = \frac12$ El ángulo cuyo coseno es $1/2$ es $60^\circ$. Como el coseno es positivo (1º y 4º cuadrante):

    Solución general para este caso: $$x = \pm 60^\circ + 360^\circ \cdot k$$

    • Solución Final: $$x = 360^\circ \cdot k$$ $$x = \pm 60^\circ + 360^\circ \cdot k$$ Solución General: $x = 2\pi n \quad ;

    To solve trigonometric equations in 1º Bachillerato, the main goal is to use identities to express the equation in terms of a single trigonometric function (like sinxsine x cosxcosine x ) and then find all possible angles that satisfy it. Fundamental Steps for Success Simplify Using Identities: Use formulas like or double-angle formulas ( ) to reduce the equation to a single reason. Factor or Change Variables: Often, you can treat sinxsine x cosxcosine x as "z" to solve it like a quadratic equation (

    Find All Solutions: Remember that trigonometric functions are periodic. A basic solution usually comes with +360∘kpositive 360 raised to the composed with power k ) to account for all laps around the circle. Exercise 1: Basic Linear Equation Solve: Isolate the Function: Find the Primary Angles:On the unit circle, the sine is 12one-half (Quadrant I) (Quadrant II) General Solution:✅ Exercise 2: Using the Pythagorean Identity Solve: Convert to a Single Function:Use Rearrange into Quadratic Form: Solve for sinxsine x :Using the quadratic formula for Final Answer:✅ The solutions are 330∘330 raised to the composed with power 360∘k360 raised to the composed with power k Exercise 3: Double Angle Equation Solve: Apply Double Angle Formula: Factor Out the Common Term: Solve Each Factor: 90∘90 raised to the composed with power 270∘270 raised to the composed with power Final Answer:✅

    For more practice, you can find categorized PDF worksheets at Matemáticas Online or detailed walkthroughs on Apuntes Marea Verde.

    Aquí tienes un reporte detallado sobre las Ecuaciones Trigonométricas para el nivel de 1º de Bachillerato, completo con la teoría necesaria y una selección de ejercicios resueltos paso a paso.

    ¿Necesitas más ejercicios? Deja en los comentarios qué tipo de ecuación se te atasca más (¿con tangente? ¿con seno y coseno mezclados?) y te ayudamos.

    Comparte este post con tu compañero de clase que aún lucha con las trigonométricas.

    Here’s a step-by-step study guide with solved exercises for trigonometric equations at the 1º Bachillerato level (ages 16–17, typically the first year of Spanish Bachillerato).
    The focus is on fixed (standard) equation types: basic, quadratic, requiring identities, and those solved by factoring.

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