QuestionSolve $\sin 2 x=\cos (x-30)$ for $-90^{\circ} \leq x<90^{\circ}$ .

Studdy Solution

STEP 1

Assumptions1. We are solving the equation \sinx = \cos(x -30^{\circ}) for $-90^{\circ} \leq x <90^{\circ}$ . . We are looking for solutions in the domain $-90^{\circ} \leq x <90^{\circ}$ .
3. We know that \sinx =\sin x\cos x (double angle formula).
4. We know that $\cos (x -30^{\circ}) = \cos x\cos30^{\circ} + \sin x\sin30^{\circ}$ (cosine of difference of two angles).

STEP 2

First, we need to rewrite the equation using the double angle formula for sine.
$\sin2x =2\sin x\cos x$

STEP 3

Next, rewrite the equation using the cosine of difference of two angles.
$\cos (x -30^{\circ}) = \cos x\cos30^{\circ} + \sin x\sin30^{\circ}$

STEP 4

Now, we can set the two equations equal to each other.
$2\sin x\cos x = \cos x\cos30^{\circ} + \sin x\sin30^{\circ}$

STEP 5

implify the equation by multiplying out the right side.
$2\sin x\cos x = \frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x$

STEP 6

Rearrange the equation to isolate terms involving $\sin x$ and $\cos x$ .
$2\sin x\cos x - \frac{\sqrt{3}}{2}\cos x = \frac{1}{2}\sin x$

STEP 7

Factor out $\cos x$ from the left side.
$\cos x(2\sin x - \frac{\sqrt{3}}{2}) = \frac{1}{2}\sin x$

STEP 8

Divide both sides by $\cos x$ to isolate $\sin x$ .
$2\sin x - \frac{\sqrt{3}}{2} = \frac{1}{2}\tan x$

STEP 9

Rearrange the equation to isolate $\tan x$ .
$\tan x =4\sin x - \sqrt{3}$

STEP 10

We know that $\tan x = \sin x / \cos x$ . Substitute this into the equation.
$\frac{\sin x}{\cos x} =4\sin x - \sqrt{3}$

STEP 11

Multiply both sides by $\cos x$ to isolate $\sin x$ .
$\sin x =4\sin x\cos x - \sqrt{3}\cos x$

STEP 12

Rearrange the equation to isolate terms involving $\sin x$ and $\cos x$ .
$\sin x( -4\cos x) = -\sqrt{}\cos x$

STEP 13

Divide both sides by $( -\cos x)$ to isolate $\sin x$ .
$\sin x = \frac{-\sqrt{3}\cos x}{ -\cos x}$

STEP 14

We know that $\sin^2 x + \cos^2 x =$ . Substitute $\sin x$ from the equation above into this equation.
$\left(\frac{-\sqrt{3}\cos x}{ -4\cos x}\right)^2 + \cos^2 x =$

STEP 15

implify the equation.
$3\cos^2 x +4\cos^4 x -4\cos^3 x + \cos^2 x - =0$

STEP 16

Rearrange the equation to isolate terms involving $\cos x$ .
$4\cos^4 x -4\cos^3 x +4\cos^2 x - =0$

STEP 17

Factor out $\cos x$ from the equation.
$\cos x(4\cos^3 x -4\cos^2 x +4\cos x -) =0$

STEP 18

Set each factor equal to zero and solve for $\cos x$ .
$\cos x =0, \quad4\cos^3 x -4\cos^2 x +4\cos x - =0$

STEP 19

olve the cubic equation for $\cos x$ .
$\cos x = \frac{1}{}, \quad \cos x = -\frac{1}{}$

STEP 20

Find the corresponding $x$ values for each solution of $\cos x$ .
$x = \arccos(0), \quad x = \arccos\left(\frac{}{}\right), \quad x = \arccos\left(-\frac{}{}\right)$

STEP 21

Calculate the $x$ values.
$x =90^{\circ}, \quad x =60^{\circ}, \quad x = -60^{\circ}$ So, the solutions to the equation \sinx = \cos(x -30^{\circ}) for $-90^{\circ} \leq x <90^{\circ}$ are $x =60^{\circ}$ and $x = -60^{\circ}$ .

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