Use your graph to determine 3 possible solutions for for
.

Ask by Hammond Daniel. in South Africa

Mar 09,2025

Question

Use your graph to determine 3 possible solutions for $$ \sin 2 x \cos x=\cos x $$ for $$ x \in\left[90^{\circ} ; 270^{\circ}\right] $$.

UpStudy AI · Accepted Answer

We start with the equation
$$
\sin 2x \cos x = \cos x.
$$
We can factor the right-hand side by writing the equation as
$$
\cos x (\sin 2x - 1) = 0.
$$
This product is zero if at least one of the factors is zero.

1. **Case 1: $$\cos x = 0$$.**

The cosine function is zero when
   $$
   x = 90^\circ,\, 270^\circ,\, 450^\circ,\dots
   $$
   Within the interval $$\left[90^\circ, 270^\circ\right]$$, we have the solutions
   $$
   x = 90^\circ \quad \text{and} \quad x = 270^\circ.
   $$

2. **Case 2: $$\sin 2x - 1 = 0$$.**

Solve for $$\sin 2x = 1$$. The sine function reaches 1 when its argument is $$90^\circ$$ plus any multiple of $$360^\circ$$:
   $$
   2x = 90^\circ + 360^\circ k,\quad k\in\mathbb{Z}.
   $$
   Solving for $$x$$ gives
   $$
   x = 45^\circ + 180^\circ k.
   $$
   We need $$x$$ to be in $$\left[90^\circ, 270^\circ\right]$$. Testing values of $$k$$:
   - For $$k = 0$$: $$x = 45^\circ$$, which is less than $$90^\circ$$.
   - For $$k = 1$$: $$x = 225^\circ$$, which lies in $$\left[90^\circ, 270^\circ\right]$$.
   - For $$k = 2$$: $$x = 405^\circ$$, which is greater than $$270^\circ$$.
   
   So the only solution from this case is
   $$
   x = 225^\circ.
   $$

Combining the results from both cases gives three solutions in the interval $$\left[90^{\circ}, 270^{\circ}\right]$$:
$$
x = 90^\circ,\quad x = 225^\circ,\quad x = 270^\circ.
$$
The solutions are $$ x = 90^\circ $$, $$ x = 225^\circ $$, and $$ x = 270^\circ $$.

Use your graph to determine 3 possible solutions for for
.

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