# Thread:

1. ## Solving Trig Equation

Hi all,

$3\sin^2\theta + 7\cos\theta = 5$ solve for $0 <= \theta <= 2\pi$

The working I've done so far.

1. $3(1 - \cos^2\theta) + 7\cos\theta = 5$

2. $3 - 3\cos^2\theta + 7\cos\theta = 5$

3. $\cos\theta (7 - 3\cos\theta) = 2$

4. $\cos\theta = 2$ or $(7 - 3\cos\theta) = 2$

5. $7 - 3\cos\theta = 2$

6. $-3\cos\theta = -5$

7. $\cos\theta = \frac{3}{5}$

Therefore, $\theta = 53.13 deg$ or $\theta = 306.87 deg$

Which means that $\theta = \frac{53.13\pi}{180}$ or $\theta = \frac{306.87\pi}{180}$

I'm just not sure if step 4 is correct. Can I simply ignore one of the options because $\cos\theta = 2$ is invalid?

Cheers.

2. Step 4 is correct.

3. Thanks How about the rest of it lol.

4. I'm sorry; step 4 is incorrect!

How does step 4 follow from step 3.

3. $\cos\theta (7 - 3\cos\theta) = 2$

How did you get to step 4?

4. $\cos\theta = 2$ or $(7 - 3\cos\theta) = 2$

This conclusion can be made only if the RHS=0. I think you're missing something.

5. I was thinking along the lines of factorising.

I.e.

$(x^2 + 3x) = 0$
$x (x + 3) = 0$
$x = 0$ or $(x + 3) = 0$
$x = 0$ or $x = -3$

Edit - Posted before your addition, I can only do that when x = 0 ? Learnt something new (although I should have known that before I guess)

Any further idea's of where to look?

6. $3cos^2 \theta - 7cos \theta+2=0$
$(3cos^2 \theta-6cos \theta - cos \theta+2)=0$
$3cos \theta (cos\theta-2) - (cos \theta-2)=0$
$(cos \theta-2) (3cos \theta-1)=0$
$(cos \theta-2)=0$ or $(3cos \theta-1)=0$

7. Originally Posted by Peleus
Hi all,

$3\sin^2\theta + 7\cos\theta = 5$ solve for $0 <= \theta <= 2\pi$

The working I've done so far.

1. $3(1 - \cos^2\theta) + 7\cos\theta = 5$

2. $3 - 3\cos^2\theta + 7\cos\theta = 5$

3. $\cos\theta (7 - 3\cos\theta) = 2$

4. $\cos\theta = 2$ or $(7 - 3\cos\theta) = 2$

5. $7 - 3\cos\theta = 2$

6. $-3\cos\theta = -5$

7. $\cos\theta = \frac{3}{5}$

Therefore, $\theta = 53.13 deg$ or $\theta = 306.87 deg$

Which means that $\theta = \frac{53.13\pi}{180}$ or $\theta = \frac{306.87\pi}{180}$

I'm just not sure if step 4 is correct. Can I simply ignore one of the options because $\cos\theta = 2$ is invalid?

Cheers.
you can't do that on step three you need to do the following:

3-3cos^2x+7cosx=5

take everything to the right hand side
3cos^2x-7cosx+5-3=0
3cos^2x-7cosx+2=0

now you have to factorise
i would replace the cos with y first
3y^2-7y+2=0
(3y-1)(y-2)=0

so
3cosx=1
cosx=1/3

or cosx=2
x=error

so x=70.53 or 289.47 (2dp)

and i have used x instead of theta
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 45, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8953985571861267, "perplexity_flag": "middle"}