Question

Solve 2sin2(x) – 5sin(x) – 3 = 0. Let u = sin(x). Which of the following is equivalent to the given equation?

Answer 1

Answer:

$x=(-1)^{k+1}\cdot \dfrac{\pi}{6}+\pi k,\ k\in Z$

Step-by-step explanation:

Consider the equation

$2\sin^2x-5\sin x-3=0$

Substitute

$u=\sin x$

into the equation:

$2u^2-5u-3=0$

This is quadratic equation, where

$D=(-5)^2-4\cdot 2\cdot (-3)=25+24=49\\ \\\sqrt{D}=7$

Hence,

$u_1=\dfrac{-(-5)-\sqrt{D}}{2\cdot 2}=\dfrac{5-7}{4}=-\dfrac{1}{2}\\ \\u_2=\dfrac{-(-5)+\sqrt{D}}{2\cdot 2}=\dfrac{5+7}{4}=3$

Now, use the substitution again:

1. When $u_1=-\dfrac{1}{2},$ then

$\sin x=-\dfrac{1}{2}\\ \\x=(-1)^k\sin^{-1}\left(-\dfrac{1}{2}\right)+\pi k,\ k\in Z\\ \\x=(-1)^k\cdot \left(-\dfrac{\pi}{6}\right)+\pi k,\ k\in Z\\ \\x=(-1)^{k+1}\cdot \dfrac{\pi}{6}+\pi k,\ k\in Z$

2. When $u_2=3,$ then

$\sin x=3$

This equation has no solutions, because the range of $\sin x$ is $[-1,1]$ and 3>1.

Answer 2

Answer:

(2u + 1)(u – 3) = 0 is equivalent to the given equation

For the second part of the question, the solutions to the equation are C and D, 7π/6 +2kπ and 11π/6 + 2kπ

Step-by-step explanation: