Maths — Pure Mathematics 3

How many solutions does $2 \cos^2x + 5\sin x = 4$ have?

Review question2.3

How many values of $x$ satisfy the equation $2 \cos^2x + 5\sin x = 4$ in the range $0 \leq x < 2 \pi$ ?

Can we rewrite the equation in terms of $\sin x$ only or $\cos x$ only?

We have

\begin{align*} & 2 \cos^2x + 5\sin x = 4 \\ \iff & 2 - 2\sin^2 x + 5 \sin x = 4 \\ \iff & 2\sin^2 x - 5 \sin x + 2 = 0 \\ \iff & (2\sin x - 1)(\sin x - 2) = 0, \end{align*}

so $\sin x = \frac{1}{2}$ (since $\sin x = 2$ is impossible).

This occurs at $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$ in the given range, so there are $2$ such values of $x$ and the answer is (a).

How many solutions does

2 \cos^2x + 5\sin x = 4

have?

Question

How many values of

x

satisfy the equation

2 \cos^2x + 5\sin x = 4

in the range

0 \leq x < 2 \pi

2

4

6

8

Suggestion

Can we rewrite the equation in terms of

\sin x

only or

\cos x

only?

Solution

We have

\begin{align*} & 2 \cos^2x + 5\sin x = 4 \\ \iff & 2 - 2\sin^2 x + 5 \sin x = 4 \\ \iff & 2\sin^2 x - 5 \sin x + 2 = 0 \\ \iff & (2\sin x - 1)(\sin x - 2) = 0, \end{align*}

\sin x = \frac{1}{2}

(since

\sin x = 2

is impossible).

This occurs at

x = \dfrac{\pi}{6}

and

x = \dfrac{5\pi}{6}

in the given range, so there are

2

such values of

x

and the answer is (a).

How many solutions does 2cos⁡2x+5sin⁡x=42 \cos^2x + 5\sin x = 42cos2x+5sinx=4 have?