Thread: Solve (3 sin x + 4 cos x)^2 = 3^2

1. Solve (3 sin x + 4 cos x)^2 = 3^2

I need to solve this equation (3 sin x + 4 cos x)^2 = 3^2.

I have solved this far, and it doesn't seem that I'm getting anywhere to the solution.

9 sin^2 x + 24 sin x cos x + 16 cos^2 x = 9
9 sin^2 x + 12 sin 2x + 16 cos^2 x = 9
9 sin^2 x + 12 sin 2x + 16(1 - sin^2 x) = 9
9 sin^2 x + 12 sin 2x + 16 - 16 sin^2 x = 9
- 7 sin^2 x + 12 sin 2x = -7

Where do I go from here?! It doesn't look I can factorise...

2. Thanks so much! Finally got this solved. So, x has two values = 90 deg, 343.7 deg (or -16.3 deg)!

3. You're welcome!

4. Originally Posted by darkvelvet
Thanks so much! Finally got this solved. So, x has two values = 90 deg, 343.7 deg (or -16.3 deg)!
$\displaystyle + 360^{\circ}n$ where $\displaystyle n$ is an integer.

5. Hello, darkvelvet!

$\displaystyle \text{Solve: }\;(3\sin x + 4\cos x)^2 \:=\: 3^2$

Take square roots: .$\displaystyle 3\sin x + 4\cos x \:=\:\pm3$

Divide by 5: . $\displaystyle \frac{3}{5}\sin x + \frac{4}{5}\cos x \;=\;\pm\frac{3}{5}$ .[1]

Let $\displaystyle \,\theta$ be an angle in a right triangle with: .$\displaystyle opp = 4,\;adj = 3,\;hyp = 5$
. . Then: .$\displaystyle \sin\theta = \frac{4}{5},\;\cos\theta = \frac{3}{5},\;\tan\theta = \frac{4}{3}$

Substitute into [1]: .$\displaystyle \cos\theta\sin x + \sin\theta\cos x \:=\:\pm\frac{3}{5}$

Then we have: . $\displaystyle \sin(x + \theta) \;=\;\pm\frac{3}{5}$

. . . . . . . . . . . . . . .$\displaystyle x + \theta \;=\;\arcsin\left(\pm\frac{3}{5}\right)$

. . . . . . . . . . . . . . . . . $\displaystyle x \;=\;\arcsin\left(\pm\frac{3}{5}\right) - \theta$

. . . . . . . . . . . . . . . . . $\displaystyle x \;=\;\arcsin\left(\pm\frac{3}{5}\right) - \arcsin\left(\frac{4}{5}\right)$