Explained: "4sin⁡(x)+7cos⁡(x)=6where 0≤x≤360∘I put the equation into the form..."

High School
Geometry
Trigonometry
Trigonometric equations and identities

aspifsGak5u

Answered question

2022-01-03

4 \sin (x) + 7 \cos (x) = 6

where

0 \leq x \leq 360^{\circ}

I put the equation into the form

a \sin (x) + b \cos (x) = R \sin (x + a)

, but after determining that

R \cos (a) = 4, R \sin (a) = 7

and

R \sin (x + a) = 6

, I don't know how to proceed.

Answer & Explanation

ramirezhereva

Beginner2022-01-04Added 28 answers

Starting from

R = \sqrt{66}, a = \arcsin \frac{7}{\sqrt{65}}

we have

\sqrt{65} \sin (x + a) = 6

\Rightarrow x = \arcsin \frac{6}{\sqrt{65}} - a = a r c i s n \frac{6}{\sqrt{65}} - \arcsin \frac{7}{\sqrt{65}}

Using

\arcsin u - \arcsin v = \arcsin (u \sqrt{1 - v^{2}} - v \sqrt{1 - u^{2}})

x = \arcsin (\frac{6}{\sqrt{65}} \cdot \frac{4}{\sqrt{65}} - \frac{7}{\sqrt{65}} \cdot \frac{\sqrt{65 - 6^{2}}}{\sqrt{65}})

x = \arcsin (\frac{24 - 7 \sqrt{29}}{65})

Vasquez

Expert2022-01-08Added 669 answers

HINT:
NSK
The R is related to
so that $\sin a = \frac{7}{\sqrt{65}}, \cos a = \frac{4}{\sqrt{65}}$
which is more convenient. Divide both sides by $\sqrt{65}$
$\sin (x + a) = \frac{6}{\sqrt{65}}$
where
$\tan α = \frac{7}{4}$

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