Math Problem Statement
Solve the given equation. (a) sin θ = √2 2 (b) √2 cos θ + 1 = 0 (rationalize to get a value we know) (c) tan θ + 1 = 0 (d) sin2 x − sin x − 2 = 0
Solution
Let us solve each equation step by step:
(a)
The value corresponds to the angles where sine equals this value:
- or in the first quadrant.
- or in the second quadrant.
Thus, the solutions are:
(b)
Step 1: Isolate :
Step 2: Rationalize:
Step 3: Solve for :
The value corresponds to the angles where cosine is negative:
- in the second quadrant.
- in the third quadrant.
Thus, the solutions are:
(c)
Step 1: Isolate :
Step 2: Solve for :
The value corresponds to the angles where tangent is negative:
- in the second quadrant.
- in the fourth quadrant.
Thus, the solutions are:
(d)
This is a quadratic equation in . Let , so the equation becomes:
Step 1: Factorize the quadratic equation:
Step 2: Solve for :
u - 2 = 0 \quad \Rightarrow \quad u = 2 \quad \text{(Not possible as \sin x \leq 1)}
Thus, .
Step 3: Solve for :
The value occurs at:
Final Solutions:
- (a)
- (b)
- (c)
- (d)
Would you like detailed steps for any specific part, or do you have any questions?
Here are 5 follow-up questions to explore:
- What are the steps to derive trigonometric values from the unit circle?
- How does the periodicity of trigonometric functions affect solutions?
- What is the significance of rationalizing terms in trigonometric equations?
- How can you graphically verify these solutions?
- How does factoring help solve trigonometric quadratic equations?
Tip: Always verify solutions in the interval you are working within, especially for periodic functions like sine and cosine!
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Math Problem Analysis
Mathematical Concepts
Trigonometry
Trigonometric Equations
Quadratic Equations
Formulas
sin(θ) = √2/2 → θ = π/4, 3π/4
cos(θ) = -1/√2 → θ = 3π/4, 5π/4
tan(θ) = -1 → θ = 3π/4, 7π/4
Factoring sin²(x) − sin(x) − 2 = 0
Theorems
Unit Circle Theorem
Periodic Properties of Trigonometric Functions
Quadratic Equation Theorem
Suitable Grade Level
Grades 10-12
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