Ecuaciones Trigonometricas 1 Bachillerato Ejercicios Resueltos Fixed ((free)) ◎

$$x = 30^\circ + 360^\circ \cdot k$$ $$x = 150^\circ + 360^\circ \cdot k$$ (Si quisiéramos la solución en el intervalo $[0, 360^\circ)$, serían $30^\circ$ y $150^\circ$).

Solve ( \tan x = \sqrt3 ) (general solution).

| Ángulo (grados) | Ángulo (radianes) | (\sin) | (\cos) | (\tan) | |----------------|-------------------|----------|----------|----------| | 0° | 0 | 0 | 1 | 0 | | 30° | (\pi/6) | 1/2 | (\sqrt3/2) | (\sqrt3/3) | | 45° | (\pi/4) | (\sqrt2/2) | (\sqrt2/2) | 1 | | 60° | (\pi/3) | (\sqrt3/2) | 1/2 | (\sqrt3) | | 90° | (\pi/2) | 1 | 0 | No definida | | 180° | (\pi) | 0 | -1 | 0 | | 270° | (3\pi/2) | -1 | 0 | No definida | | 360° | (2\pi) | 0 | 1 | 0 |

, etc.) donde la incógnita es un ángulo. A diferencia de las ecuaciones algebraicas, estas suelen tener debido a la periodicidad de las funciones trigonométricas.

En esta ecuación tenemos senos y cosenos mezclados. Para resolverla, debemos unificar todo a una sola razón trigonométrica. Usamos la identidad fundamental despejando el coseno: Sustituimos en la ecuación:

$$(1 - \sin^2 x) - \sin x = 1$$

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: ✅

If ( \tan x = \tan \alpha ), then: [ x = \alpha + k\pi,\ k \in \mathbbZ ]

cos(x)⋅(2sin(x)−1)=0cosine x center dot open paren 2 sine x minus 1 close paren equals 0 Caso A: Caso B: . Esto ocurre en: Ejercicio Resuelto 2: Cambio de Variable Enunciado: Resuelve Uniformar Razones: Cambiamos para tener todo en función del seno.