Chapter 12: Q. 10 (page 840)

Solve the equation $2 \sin^{2} x - \sin x - 3 = 0, 0 \leq x < 2 π$

Short Answer

Expert verified

The solution of the equation is $\frac{2 π}{3} .$

Step by step solution

Step 1. Given Information

Given trigonometric equation is $2 \sin^{2} x - \sin x - 3 = 0, 0 \leq x < 2 π$ .

Let $\sin x = y,$ then $2 y^{2} - y - 3 = 0$

Step 2. Solving Equation

$2 y^{2} - y - 3 = 0 2 y^{2} - 3 y + 2 y - 3 = 0 y (2 y - 3) + 1 (2 y - 3) = 0 (y + 1) (2 y - 3) = 0 y = - 1, \frac{3}{2}$

Step 3. Finding solution for sinx

We get $\sin x = - 1, \sin x = \frac{3}{2},$

role="math" localid="1646989043028" $x = π + \frac{π}{2}, 2 π - \frac{π}{2} x = \frac{3 π}{2},$

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Solve the equation $2 \sin^{2} x - \sin x - 3 = 0, 0 \leq x < 2 π$

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solving Equation

Step 3. Finding solution for sinx

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Solve the equation2sin2x-sinx-3=0,0≤x<2π

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solving Equation

Step 3. Finding solution for sinx

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Statistics

Applied Mathematics

Theoretical and Mathematical Physics

Logic and Functions

Decision Maths

Study anywhere. Anytime. Across all devices.

Solve the equation $2 \sin^{2} x - \sin x - 3 = 0, 0 \leq x < 2 π$