Question

 Use trigonometric identities to solve each equation within the given domain

sec(x) cos(3x) = 0 from [- x /2 , x /2]

Answer 1

Answer:

Either $x = (-\pi / 6)$ or $x = (\pi / 6)$.

Step-by-step explanation:

Angle sum identity for cosine:

$\cos(a + b) = \cos(a)\, \cos(b) - \sin(a) \, \sin(b)$.

Double-angle identity for cosine and sine:

$\begin{aligned}& \cos(2\, a)\\ &= \cos^{2}(a) - \sin^{2} (a) \\ &= (\cos^{2}(a) + \sin^{2}(a) - 2\, \sin^{2}(a) \\ &= 1 - 2\, \sin^{2}(a)\end{aligned}$.

$\sin(2\, a) = 2\, \cos(a) \, \sin(a)$.

The secant of an angle is the inverse reciprocal of the cosine of that angle:

$\displaystyle \text{ \sec(x) = \frac{1}{\cos(x)} if \cos(x) \ne 0 }$.

Apply these identities to rewrite the left-hand side of the equation.

$\begin{aligned} & \sec(x) \, \cos(3\, x) \\ =\; & \frac{1}{\cos(x)} \cdot \cos(3\, x) \\ =\; & \frac{1}{\cos(x)} \cdot (\cos(x)\, \cos(2\, x) - \sin(x)\, \sin(2\, x)) \\ =\; &\cos(2\, x) \\ & - \frac{1}{\cos(x)} \cdot (\sin(x)\, \sin(2\, x)) \\ =\; & \cos(2\, x)\\ & - \frac{1}{\cos(x)} \cdot (\sin(x) \cdot (2\, \sin(x)\, \cos(x))) \\ =\; & \cos(2\, x) - 2\, \sin^{2}(x) \\ =\; & (1 - 2\, \sin^{2}(x)) - 2\, \sin^{2}(x) \\ =\; &1 - 4\, \sin^{2}(x) \end{aligned}$.

Solve the rewritten equation:

$\displaystyle \sin^{2}(x) = \frac{1}{4}$.

Either $\sin(x) = (1/2)$ or $\sin(x) = (-1/2)$.

The question required that $(-\pi / 2) \le x \le (\pi / 2)$. The sine function is one-to-one over this range. In other words, for each $y$ between $(-1)$ and $1$, there is exactly one $x$ between $(-\pi / 2)$ and $(\pi / 2)$ such that $\sin(x) = y$.

If $\sin(x) = (-1/2)$ and $(-\pi / 2) \le x \le (\pi / 2)$, then $x = (-\pi / 6)$.

Otherwise, if $\sin(x) = (1/2)$ and $(-\pi / 2) \le x \le (\pi / 2)$, then $x = (\pi / 6)$,

Verify that $\cos(x) \ne 0$ for either of the two $x$ values, such that the $\sec(x)$ in the original equation takes a real value.