Question

Solve for real numbers:

$\sf \left|\begin{array}{cccc}1&3+sin \ x&2+3 \ sin \ x&2 \ sinx\\1&2 + sin \ x + cos \ x&2sin \ x + sin \ x \ cos \ x&sin \ 2x \\1&1+sin \ x+cos \ x&sin \ x + cos \ x + sinx \ cosx&sinx \ cosx\\1&3+ cos \ x&2+3 \ cosx& 2 \ cos x\end{array}\right| = 0$

Answer 1

solution is attached above .....

Answer 2

Step-by-step explanation:

Explanation:

sin

x

+

cos

x

=

2

3

Let's first use linear combination of cosine and sine with equal arguments formula to simplify it.

That is, we want to express

A cos x +B sin x in the form C cos (x-D). Note that A is the coefficient of cos x and B is the coefficient of sine x.

To find C use pythagorean formula and to find D we use one of these two formulas

cos

D

=

A

C

,

sin

D

=

B

C

From the given equation A = 1 and B = 1. So let's find C using pythagorean theorem.

C

=

√

A

2

+

B

2

=

√

1

2

+

1

2

=

√

2

To find D we need to first figure out which quadrant x is in and because both cos x and sin x are positive it means that x is in quadrant